Transcript
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge.
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”.
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810. Expressions calculate future results.
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810. Expressions calculate future results. Now if we know the cost is $810 but forget how many pizzas were ordered we may retrace to find x:
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810. Expressions calculate future results. Now if we know the cost is $810 but forget how many pizzas were ordered we may retrace to find x:
8x + 10 = 810
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810. Expressions calculate future results. Now if we know the cost is $810 but forget how many pizzas were ordered we may retrace to find x:
8x + 10 = 810
8x = 800–10 –10 subtract the delivery cost,
so the pizzas cost $800,
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810. Expressions calculate future results. Now if we know the cost is $810 but forget how many pizzas were ordered we may retrace to find x:
8x + 10 = 810
8x = 800–10 –10 subtract the delivery cost,
so the pizzas cost $800, divide this by $8/per pizza,
x = 100 to recover x = 100 pizzas8 8
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810. Expressions calculate future results. Now if we know the cost is $810 but forget how many pizzas were ordered we may retrace to find x:
8x + 10 = 810
8x = 800–10 –10 subtract the delivery cost,
so the pizzas cost $800, divide this by $8/per pizza,
x = 100 to recover x = 100 pizzas8 8
“8x + 10 = 810” is called an equation.
Expressions
Recall the prices for the pizzas from Pizza Grande. Each pizza is $8 and there is $10 delivery charge. Hence if we want x pizzas delivered, then the total cost is given by the expression “8x + 10”. If we order x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
Expressions calculate future results. Now if we know the cost is $810 but forget how many pizzas were ordered we may retrace to find x:
8x + 10 = 810
8x = 800–10 –10 subtract the delivery cost,
so the pizzas cost $800, divide this by $8/per pizza,
x = 100 to recover x = 100 pizzas8 8
“8x + 10 = 810” is called an equation. We want to solve equations, i.e. to backtrack from known results to find the original input x or x’s.
Solving Equations
Two expressions set equal to each other is called an equation.
Solving Equations
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true.
Solving Equations
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial / rational / algebraic equations respectively.
Solving Equations
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial / rational / algebraic equations respectively.
Solving Equations
Example A. 3x2 – 2x = 8 We solve polynomial equations by factoring.
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial / rational / algebraic equations respectively.
Solving Equations
Example A. 3x2 – 2x = 8 We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial / rational / algebraic equations respectively.
Solving Equations
Example A. 3x2 – 2x = 8 We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this(3x + 4)(x – 2) = 0
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial / rational / algebraic equations respectively.
Solving Equations
Example A. 3x2 – 2x = 8 We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this(3x + 4)(x – 2) = 0 extract answers
x = –4/3, 2
Two expressions set equal to each other is called an equation. To solve an equation means to find value(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial / rational / algebraic equations respectively.
Solving Equations
Example A. 3x2 – 2x = 8 We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this(3x + 4)(x – 2) = 0 extract answers
x = –4/3, 2To solve other equations such as rational equations, we have to transform them into polynomial equations.
Rational EquationsSolve rational equations by clearing all denominators using the LCD.
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]x – 22 = x + 1
4+ 1
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]x – 22 = x + 1
4+ 1
(x + 1)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]x – 22 = x + 1
4+ 1
(x + 1) (x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]x – 22 = x + 1
4+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)x – 2
2 = x + 14
+ 1(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)2x + 2 = 4x – 8 + x2 – x – 2
x – 22 = x + 1
4+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)2x + 2 = 4x – 8 + x2 – x – 22x + 2 = x2 + 3x – 10
x – 22 = x + 1
4+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)2x + 2 = 4x – 8 + x2 – x – 22x + 2 = x2 + 3x – 10 0 = x2 + x – 12
x – 22 = x + 1
4+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)2x + 2 = 4x – 8 + x2 – x – 22x + 2 = x2 + 3x – 10 0 = x2 + x – 12 0 = (x + 4)(x – 3) x = -4, 3
x – 22 = x + 1
4+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Rational EquationsSolve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
(x – 2)(x + 1) * [ ]2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)2x + 2 = 4x – 8 + x2 – x – 22x + 2 = x2 + 3x – 10 0 = x2 + x – 12 0 = (x + 4)(x – 3) x = -4, 3 Both are good.
x – 22 = x + 1
4+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 22 = x + 1
4 + 1
Factoring By GroupingSome polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
2x3 – 3x2 – 8x + 12 = 0
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0 (2x – 3)(x – 2)(x + 2) = 0
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
So x = 2/3, 2, –2
2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0 (2x – 3)(x – 2)(x + 2) = 0
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
So x = 2/3, 2, –2
2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0 (2x – 3)(x – 2)(x + 2) = 0
Except for some special cases, polynomial equations with degree 3 or more are solved with computers.
Factoring By Grouping
Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out common factors twice. We call this factor by grouping.
So x = 2/3, 2, –2
2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0 (2x – 3)(x – 2)(x + 2) = 0
We may also use the quadratic formula to solve all2nd degree polynomial equations.
Except for some special cases, polynomial equations with degree 3 or more are solved with computers.
Factoring By Grouping
Quadratic Formula and DiscriminantQuadratic Formula (QF):
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
x =–b ± b2 – 4ac
2aA “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.
x =–b ± b2 – 4ac
2aA “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots,
x =–b ± b2 – 4ac
2aA “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots, furthermore if b2 – 4ac is a perfect square, the roots are rational.
x =–b ± b2 – 4ac
2aA “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots, furthermore if b2 – 4ac is a perfect square, the roots are rational.II. If b2 – 4ac < 0, the roots are not real.
x =–b ± b2 – 4ac
2aA “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots, furthermore if b2 – 4ac is a perfect square, the roots are rational.II. If b2 – 4ac < 0, the roots are not real.
x =–b ± b2 – 4ac
2a
Example D. Find the values of k where the solutions are real for x2 + 2x + (2 – 3k) = 0
A “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots, furthermore if b2 – 4ac is a perfect square, the roots are rational.II. If b2 – 4ac < 0, the roots are not real.
x =–b ± b2 – 4ac
2a
Example D. Find the values of k where the solutions are real for x2 + 2x + (2 – 3k) = 0 We need b2 – 4ac > 0,
A “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots, furthermore if b2 – 4ac is a perfect square, the roots are rational.II. If b2 – 4ac < 0, the roots are not real.
x =–b ± b2 – 4ac
2a
Example D. Find the values of k where the solutions are real for x2 + 2x + (2 – 3k) = 0 We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
A “root” is a solution for the equation “# = 0”.
Quadratic Formula and DiscriminantQuadratic Formula (QF):The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value indicates the type of roots we have.I. If b2 – 4ac > 0, we have real roots, furthermore if b2 – 4ac is a perfect square, the roots are rational.II. If b2 – 4ac < 0, the roots are not real.
x =–b ± b2 – 4ac
2a
Example D. Find the values of k where the solutions are real for x2 + 2x + (2 – 3k) = 0 We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
– 4 + 12k > 0 or k > 1/3
A “root” is a solution for the equation “# = 0”.
Power EquationsEquations of the Form xp/q = c
Power EquationsThe solution to the equation x 3 = –8
Equations of the Form xp/q = c
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Equations of the Form xp/q = c
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8
Equations of the Form xp/q = c
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8 then
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8 then
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2 Rational power equations are equations of the type xR = c where R = p/q is a rational number.
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8 then
To solve a power equation, take the reciprocal power,
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2 Rational power equations are equations of the type xR = c where R = p/q is a rational number.
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8 then
To solve a power equation, take the reciprocal power, so if xR = c,
The reciprocal of the power 3
xp/q = c
Equations of the Form xp/q = c
x = (–8)1/3 = –2 Rational power equations are equations of the type xR = c where R = p/q is a rational number.
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8 then
To solve a power equation, take the reciprocal power, so if xR = c,
then x = (±)c1/R
The reciprocal of the power 3
xp/q = c
Equations of the Form xp/q = c
x = (–8)1/3 = –2 Rational power equations are equations of the type xR = c where R = p/q is a rational number.
Reciprocate the powers
Power EquationsThe solution to the equation x 3 = –8 is x = √–8 = –2. 3
Using fractional exponents, we write these steps asif x3 = –8 then
To solve a power equation, take the reciprocal power, so if xR = c,
then x = (±)c1/R
Reciprocate the powers
The reciprocal of the power 3
xp/q = c
x = (±)cQ/Por
Equations of the Form xp/q = c
x = (–8)1/3 = –2 Rational power equations are equations of the type xR = c where R = p/q is a rational number.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16 Raise both sides to -3/2 power.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
Raise both sides to -3/2 power.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution.
Raise both sides to -3/2 power.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution. Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution. Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution. Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution. Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution. Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4 2x = 7 x = 7/2
Equations of the Form xp/q = cSolve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
x= 1/64 and it's a solution. Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4 2x = 7 x = 7/2
Since x = 7/2 doesn't work because 43/2 = -8,there is no solution.
Radical Equations
Radical EquationsSolve radical equations by squaring both sides to remove the square root.
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12 divide by 4; 2x = x – 3
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12 divide by 4; 2x = x – 3 square again; ( 2x)2 = (x – 3)2
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12 divide by 4; 2x = x – 3 square again; ( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12 divide by 4; 2x = x – 3 square again; ( 2x)2 = (x – 3)2
4x = x2 – 6x + 9 0 = x2 – 10x + 9
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12 divide by 4; 2x = x – 3 square again; ( 2x)2 = (x – 3)2
4x = x2 – 6x + 9 0 = x2 – 10x + 9 0 = (x – 9)(x – 1) x = 9, x = 1
Radical EquationsSolve radical equations by squaring both sides to remove the square root. Do it again if necessary.Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4 isolate the radical; 8x = 4x – 12 divide by 4; 2x = x – 3 square again; ( 2x)2 = (x – 3)2
4x = x2 – 6x + 9 0 = x2 – 10x + 9 0 = (x – 9)(x – 1) x = 9, x = 1 Only 9 is good.
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|.
Absolute Value Equations
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=x if x is positive or 0.
–x (opposite of x) if x is negative.{
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=x if x is positive or 0.
–x (opposite of x) if x is negative.{Hence | -5 | = –(-5) = 5.
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=x if x is positive or 0.
–x (opposite of x) if x is negative.{Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as |x4 – 3x + 1 | = – 2 doesn't have any solution.
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=x if x is positive or 0.
–x (opposite of x) if x is negative.{Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as |x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=x if x is positive or 0.
–x (opposite of x) if x is negative.{Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as |x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.Warning: In general |x ± y| |x| ± |y|.
The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=x if x is positive or 0.
–x (opposite of x) if x is negative.{Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as |x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.Warning: In general |x ± y| |x| ± |y|.For instance, | 2 – 3 | |2| – |3| |2| + |3|.
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a.
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Example H. a. If | x | = 3
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Example H. a. If | x | = 3 then x = 3 or x= –3
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Also if |x| = |y| then x = y or x = –y.Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Also if |x| = |y| then x = y or x = –y.Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5c. Solve |2x – 3| = |3x + 1|
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Also if |x| = |y| then x = y or x = –y.Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Also if |x| = |y| then x = y or x = –y.Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1) 4 = x or 2x – 3 = –3x – 1
Because |x±y| |x|±|y|, we have to solve absolute value equations by rewriting it into two equations without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Also if |x| = |y| then x = y or x = –y.Example H. a. If | x | = 3 then x = 3 or x= –3b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1) 4 = x or 2x – 3 = –3x – 1 4 = x or x = 2/5
Absolute Value Equations
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