124 An equation involving a variable of degree 2 is quadratic equation. a a a a m m (m+2) (m+2) Consider a square of side ‘a’ units and its area 25 square units. Area of the square = (side) 2 25 = a 2 or a 2 = 25 .............. (2) In equation (2) the degree of the variable is two What do you call such an equation? Such an equation is a quadratic equation. a 2 = 25 ∴ a = ± 5 a = + 5 or a = –5 Consider a rectangle of sides ‘m’ and ‘(m + 2)’ units and its area is 8 sq units. Area of a rectangle = (length) (breadth) 8 = (m) (m + 2) 8 = m 2 + 2m or m 2 + 2m = 8 .......................... (3) Compare the equation (2) and (3) In equation (2) a 2 = 25, variable occurs only in second degree. In equation (3) m 2 + 2m = 8, variable occurs in second degree as well as in first degree. Quadratic equation involving a variable only in second degree is a “Pure Quadratic Equation’’. Example : (1) x 2 =9 (2) 2a 2 = 18 If the terms in the RHS are transposed to LHS then, (1) x 2 –9=0 (2) 2a 2 – 18 = 0 An equation that can be expressed in the form ax 2 + c = 0, where a and c are real numbers and a ≠ 0 is a pure quadratic equation. A Quadratic equation has only two roots.
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124
An equation involving avariable of degree 2 isquadratic equation.
a
a
aa
m m
(m+2)
(m+2)
Consider a square of side ‘a’ units and its area 25 square units.
Area of the square = (side)2
25 = a2
or
a2 = 25 .............. (2)
In equation (2) the degree of the variable is two
What do you call such an equation?Such an equation is a quadratic equation.
a2 = 25
∴ a = ± 5
a = + 5 or a = –5
Consider a rectangle of sides ‘m’ and ‘(m + 2)’ units and its area is 8 sq units.
Area of a rectangle = (length) (breadth)
8 = (m) (m + 2)
8 = m2 + 2m
or
m2 + 2m = 8 .......................... (3)
Compare the equation (2) and (3)
In equation (2) a2 = 25, variable occurs only in second degree.
In equation (3) m2 + 2m = 8, variable occurs in second degree as well as in first degree.
Quadratic equation involving a variable only in second degree is a“Pure Quadratic Equation’’.
Example :
(1) x2 = 9 (2) 2a2 = 18
If the terms in the RHS are transposed to LHS then,
(1) x2 – 9 = 0 (2) 2a2 – 18 = 0
An equation that can be expressed in the form ax2 + c = 0, where a andc are real numbers and a ≠ 0 is a pure quadratic equation.
A Quadratic equation hasonly two roots.
125
Quadratic equation involving a variable in second degree as well as in firstdegree is an “Adfected Quadratic Equation”
Example :
(1) x2 + 3x = 10 (2) 3a2 – a = 2
If the terms in the RHS are transposed to LHS then,
(1) x2 + 3x – 10 = 0 (2) 3a2 – a – 2 = 0
ax2 + bx + c = 0 is the standard form of a quadratic equation where a, band c and variables and a ≠ 0.
1. Solving Pure Quadratic equationExample 1 : Solve the equation 3x2 – 27 = 0
Solution : 3x2 – 27 = 0
∴ 3x2 = 27
x� = 3
27
∴ x2 = 9
x = 9±x = +3 or x = –3
Example 2 : Solve the equation 4y2 – 9 = 0
Solution : 4y2 – 9 = 0∴ 4y2 = 9
∴ y2 = 4
9
y = 4
9±
y = 2
3±
y = 2
3+ or y = 2
3−
126
Example 3 : Solve the equation 99 = 4r2 – 1
Solution : 99 = 4r2 – 1
4r2 – 1 = 99
∴ 4r2 = 99 + 1
4r2 = 100
r2 = 4
100 = 25
∴ r = 25±
r = 5±r = +5 or r = –5
Example 4 : Solve the equation (m + 8)2 –5 = 31
Solution : (m + 8)2 –5 = 31
∴ (m + 8)2 = 31 + 5
(m + 8)2 = 36
∴ (m + 8)2 = 36
(m + 8) = 36± ∴ m = –8 ± 6
m = –8 + 6 or m = –8 – 6
m = –2 or m = – 14
Example 5 : If A = 2r� ; Solve for ‘r’.
Solution : A = 2rπ
2rπ = A
r2 = πA
r = π
± A
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l2
Example 6 : If l2 = r2 + h2. Solve for h and find the value of ‘h’ if l = 15 and r = 9.
Solution : l2 = r2 + h2
or
∴ r2 + h2 = l2
∴ h2 = l2 – r2
∴ h = 22 rl −±
h = 22 915 −± (substituting l = 15, r = 9)
h = 81225 −±
h = 144± h = 12±∴ h = +12 or h = –12
Example 7 : If B = 4
a.3 2
Solve for ‘a’ and find the value of ‘a’ if B = 16 3 .
Solution : B = 4
a.3 2
∴ a2 = 3
B4
a = 3
B4± (Substituting B = 16 3 )
a = 3
3164
//×±
∴ a = 64± , a = ± 8
∴ a = + 8 or a = – 8
Exercise : 5.1
A. Classify the following equations into pure and adfected quadratic equation.
1) x2 + 2 = 6 2) a2 + 3 = 2a 3) p (p – 3) = 1
4) 2m2 = 72 5) k2 – k = 0 6) 7y = y
35
128
If mn = 0, then eitherm = 0 or n = 0
B. Solve the equations
1) 5x2 = 125 2) m2 – 1 = 143 3) 4a = a
81
4) 2
2x –
4
3 =
4
17 5) (2m – 5)2 = 81 6)
18
)4( 2−x =
9
2
C.
1) If A = 2 2rπ Solve for ‘r’ and find the value of ‘r’ if A = 77 and π = 7
22
2) If V = hr2π Solve for ‘r’ and find the value of ‘r’ if V = 176 and h = 14
3) If r2 = l2 + d2 Solve for ‘d’ and find the value of ‘d’ if r = 5 and l = 4.
4) If c2 = a2 + b2 Solve for ‘b’. If a = 8 and c = 17 and find the value of ‘b’.
5) If K = 1/2mv2 Solve for ‘v’ and find the value of ‘v’ if K = 100 and m = 2
6) If v2 = u2 + 2as. Solve for ‘v’. If u = 0, a = 2 and s = 100, find the valueof v.
2. Solving the adfected quadratic equation by factorization :Example 1 : Solve the quadratic equation a2 – 3a + 2 = 0
Solution : a2 – 3a + 2 = 0
i. Resolve the expression a2 – 2a – 1a + 2 = 0
ii. Factorize a(a – 2) –1 (a – 2) = 0
iii. Taking the common factor (a – 2) (a – 1) = 0
iv. Equate each factor to zero a – 2 = 0 or a – 1 = 0
v. The roots are a = 2 or a = 1
Example 2 : Solve the quadratic equation m2 – m = 6Solution : m2 – m = 6
5. To solve the problems based on Quadratic EquationExample 1 : If the square of a number is added to 3 times the number, the sum
is 28. Find the number.
Solution : Let the number be = x
Square of the number = x2
3 times the number = 3x
Square of a number + 3 times the number = 28
x2 + 3x = 28
x2 + 3x – 28 = 0
∴ x2 + 7x – 4x – 28 = 0
x(x + 7) –4 (x + 7) = 0
(x + 7) (x – 4) = 0
x + 7 = 0 or x – 4 = 0
x = –7 or x = 4
∴ The required number is 4 or –7
138
Example 2 : Sum of a number and its reciprocal is 55
1. Find the number.
Solution : Let the number be = y
Reciprocal of the number = y
1
(Number) + (its reciprocal) = 55
1
y + y
1 =
5
26
y
1y2 + =
5
26
5(y2 + 1) = 26y
5y2 + 5 = 26y
5y2 – 26y + 5 = 0
5y2 – 25y – 1y + 5 = 0
5y (y – 5) –1 (y – 5) = 0
(y – 5) (5y – 1) = 0
Either (y – 5) = 0 or (5y – 1) = 0
y = 5 or y = 5
1
∴ The required number is 5 or 5
1
Example 3 : The base of a triangle is 4 cms longer than its altitude. If the area ofthe traingle is 48 sq cms. Find the base and altitude.
Solution : Let the altitude = x cms.Base of the triangle = (x + 4) cms.
Area of the triangle = 2
1 (base) (height)
48 = 2
1 (x + 4)x
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9) A dealer sells an article for Rs. 24 and gains as much percent as the cost priceof the article. Find the Cost price of the article.
10) Sowmya takes 6 days less than the number of days taken by Bhagya to completea piece of work. If both Sowmya and Bhagya together can complete the samework in 4 days. In how many days will Bhagya complete the work?
6. Nature of the roots of a quadratic equation.1) Consider the equation x2 – 2x + 1 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 1, b = –2, c = 1
x = a2
ac4bb 2 −±−
x = 1x2
1x1.4)2()2( 2 −−+−−
x = 2
442 −±
x = 2
02 +
x = 2
02 +or x =
2
02 −
x = 1 or x = 1 → roots are equal
2) Consider the equation x2 – 2x – 3 = 0
This is in the form ax2 + bx + c = 0
the coefficients are a = 1, b = –2, c = –3
x = a2
ac4bb 2 −±−
x = 1x2
16)2( ±−−
x = 2
42 ±+
143
x = 2
42 +or x =
2
42 −
x = 2
6or x =
2
2−
x = 3 or x = –1 → roots are distinct
3) Consider the equation x2 – 2x + 3 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 1, b = –2, c = 3
x = a2
ac4bb 2 −±−
x = 1x2
)3)(1(4)2()2( 22 −−±−−
x = 2
1242 −±
x = 2
82 −±
x = 2
222 −±
x = ( )
2
212 −± = 21 −±
x = 21 −+ or 21 −− → roots are imaginary
From the above examples it is clear that,1) Nature of the roots of quadratic equation depends upon the value of (b2 – 4ac)
2) The Expression (b2 – 4ac) is denoted by ∆ (delta) which determines the natureof the roots.
3) In the equation ax2 + bx + c = 0 the expression (b2 – 4ac) is called the discriminant.
144
Discriminant (b2 – 4ac) Nature of the roots
∆ = 0 Roots are real and equal
∆ > 0 (Positive) Roots are real and distinct
∆ < 0 (negative) Roots are imaginary
Example 1 : Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0.Consider the equation 2x2 – 5x – 1 = 0This is in form of ax2 + bx + c = 0The co-efficient are a = 2, b = –5, c = –1
∆ = b2 – 4ac∆ = (–5)2 –4(2) (–1)∆ = 25 + 8∆ = 33
∴ ∆ > 0Roots are real and distinct
Example 2 : Determine the nature of the roots of the equation 4x2 – 4x + 1 = 0Consider the equation 4x2 – 4x + 1 = 0This is in the form of ax2 + bx + c = 0The co-efficient are a = 4, b = –4, c = 1
∆ = b2 – 4ac∆ = (–4)2 –4 (4) (1)∆ = 16 – 16
∴ ∆ = 0 Roots are real and equal
Example 3 : For what values of ‘m’ roots of the equation x2 + mx + 4 = 0 are(i) equal (ii) distinctConsider the equation x2 + mx + 4 = 0This is in the form ax2 + bx + c = 0the co-efficients are a = 1, b = m, c = 4
∆ = b2 – 4ac∆ = m2 – 4(1) (4)∆ = m2 – 16
1) If roots are equal ∆ = 0 ∴ m2 – 16 = 0
m2 = 16
∴ m = 16 ∴ m = ± 4
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2) If roots are distinct ∆ > 0 ∴ m2 – 16 > 0 ∴ m2 > 16
m2 > 16m > ± 4
Example 4 : Determine the value of ‘k’ for which the equation kx2 + 6x + 1 = 0 hasequal roots.
Consider the equation kx2 + 6x + 1 = 0
This is in the form ax2 + bx + c = 0
the co-efficients are a = k, b = 6, c = 1
∆ = b2 – 4ac
since the roots are equal, b2 – 4ac = 0 (∴ ∆ = 0)
(6)2 – 4(k)(1) = 0
36 – 4k = 0
4k = 36
k = 4
36 = 9
∴ k = 9
Example 5 : Find the value of ‘p’ for which the equation x2 – (p + 2) x + 4 = 0 hasequal roots.
Consider the equation x2 – (p + 2) x + 4 = 0
This is in the form ax2 + bx + c = 0
Coefficients are a = 1, b = –(p + 2), c = 4
since the roots are equal ∆ = 0b2 – 4ac = 0
[–(p + 2)]2 – 4(1)(4) = 0
(p + 2)2 – 16 = 0
p + 2 = ± 16
p + 2 = ± 4
p + 2 = + 4 or p + 2 = –4
∴ p = 4 – 2 or p = –4 – 2
∴ p = 2 or p = –6
146
If m and n are the roots of thequadratic equation
ax2 + bx + c = 0
Sum of the roots a
b−=
Product of roots a
c+=
Exercise : 5.6
A. Discuss the nature of roots of the following equations