1.3 solving equations

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


8x + 10 = 810

Expressions


8x + 10 = 810

8x = 800–10 –10 subtract the delivery cost,

so the pizzas cost $800,

Expressions


8x + 10 = 810


so the pizzas cost $800, divide this by $8/per pizza,

x = 100 to recover x = 100 pizzas8 8

Expressions


8x + 10 = 810




“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 + 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


Solving Equations

Example A. 3x2 – 2x = 8 We solve polynomial equations by factoring.


Solving Equations


Set one side to 0, 3x2 – 2x – 8 = 0


Solving Equations


Set one side to 0, 3x2 – 2x – 8 = 0 factor this(3x + 4)(x – 2) = 0


Solving Equations


Set one side to 0, 3x2 – 2x – 8 = 0 factor this(3x + 4)(x – 2) = 0 extract answers

x = –4/3, 2


Solving Equations


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.


Example B. Solve x – 22 = x + 1

4 + 1


Example B. Solve

LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:

x – 22 = x + 1

4 + 1


Example B. Solve


(x – 2)(x + 1) * [ ]x – 22 = x + 1

4+ 1

x – 22 = x + 1

4 + 1


Example B. Solve


(x – 2)(x + 1) * [ ]x – 22 = x + 1

4+ 1

(x + 1)

x – 22 = x + 1

4 + 1


Example B. Solve


(x – 2)(x + 1) * [ ]x – 22 = x + 1

4+ 1

(x + 1) (x – 2)

x – 22 = x + 1

4 + 1


Example B. Solve


(x – 2)(x + 1) * [ ]x – 22 = x + 1

4+ 1

(x + 1) (x – 2) (x + 1)(x – 2)

x – 22 = x + 1

4 + 1


Example B. Solve


(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


Example B. Solve


(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


Example B. Solve


(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


Example B. Solve


(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


Example B. Solve


(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


Example B. Solve


(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



2x3 – 3x2 – 8x + 12 = 0




2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0




2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0




2x3 – 3x2 – 8x + 12 = 0 x2(2x – 3) – 4(2x – 3) = 0 (2x – 3)(x2 – 4) = 0 (2x – 3)(x – 2)(x + 2) = 0




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




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.




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.


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”.


b2 – 4ac is the discriminant because its value indicates the type of roots we have.

x =–b ± b2 – 4ac



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



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



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




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”.



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,




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.




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


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



Using fractional exponents, we write these steps asif x3 = –8



Using fractional exponents, we write these steps asif x3 = –8 then

The reciprocal of the power 3


x = (–8)1/3 = –2





x = (–8)1/3 = –2 Rational power equations are equations of the type xR = c where R = p/q is a rational number.



To solve a power equation, take the reciprocal power,






To solve a power equation, take the reciprocal power, so if xR = c,


xp/q = c






then x = (±)c1/R


xp/q = c



Reciprocate the powers




then x = (±)c1/R

Reciprocate the powers


xp/q = c

x = (±)cQ/Por



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.


x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2

Raise both sides to -3/2 power.


x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2

x= 1/64 and it's a solution.



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



x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2



Raise both sides to 2/3 power.


x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2



Raise both sides to 2/3 power.(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3


x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2




(2x – 3) = 4


x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2




(2x – 3) = 4 2x = 7 x = 7/2


x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2




(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


Example G. Solve x + 4 = 5x + 4


Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2


Example G. Solve x + 4 = 5x + 4 square both sides; (x + 4)2 = (5x + 4 )2 x + 2 * 4 x + 16 = 5x + 4


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


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


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



4x = x2 – 6x + 9 0 = x2 – 10x + 9



4x = x2 – 6x + 9 0 = x2 – 10x + 9 0 = (x – 9)(x – 1) x = 9, x = 1



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.




Algebraic definition of absolute value:

|x|=x if x is positive or 0.

–x (opposite of x) if x is negative.{





–x (opposite of x) if x is negative.{Hence | -5 | = –(-5) = 5.





–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.






Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.Warning: In general |x ± y| |x| ± |y|.






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.




Fact II: If |x| = a where a is a positive number, then x = a or x = –a.



Fact II: If |x| = a where a is a positive number, then x = a or x = –a. Example H. a. If | x | = 3



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



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



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



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




2x – 3 = 5 or 2x – 3 = –5c. Solve |2x – 3| = |3x + 1|




2x – 3 = 5 or 2x – 3 = –5c. Solve |2x – 3| = |3x + 1|

Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)




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




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


1.3 solving equations

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