The quadratic equation be in any of the forms given below.
to find zeroes of the polynomial, we have to replace y by 0.
A real number k of a quadratic polynomial p(x) is 0 if
p(k) = 0
To solve the quadratic equation, we can use one of the following methods.
(ii) Formula
Find the zeroes of quadratic polynomial given below.
Problem 1 :
y = (x - 4)^{2} - 25
Solution :
y= (x - 4)^{2} - 25
(x - 4)^{2 }= 25
x - 4 = √25
x - 4 = ±5
x - 4 = 5 x = 5 + 4 x = 9 |
x - 4 = -5 x = -5 + 4 x = -1 |
So, the zeroes are 9 and -1.
Problem 2 :
y = 2x^{2} - 9x - 5
Solution :
y = 2x^{2} - 9x - 5
To find zeroes, put y = 0
0 = 2x^{2} - 10x + 1x - 5
By using grouping method.
0 = 2x(x - 5) + 1(x - 5)
0 = (2x + 1) (x - 5)
Equating each factor to 0, we get
2x + 1 = 0 2x = -1 x = -1/2 |
x - 5 = 0 x = 5 |
So, the zeroes are -1/2 and 5.
Problem 3 :
f(x) = 3x^{2} + 5x - 12
Solution :
To find zeroes, Put f(x) = 0
3x^{2} + 5x - 12 = 0
3x^{2} + 9x - 4x - 12 = 0
3x(x + 3) - 4(x + 3) = 0
(3x - 4)(x + 3) = 0
Equating each factor to 0.
3x - 4 = 0 3x = 4 x = 4/3 |
x + 3 = 0 x = -3 |
So, the zeroes are -3 and 4/3.
Problem 4 :
f(x) = (-1/2) (x + 3)^{2} + 8
Solution :
To find zeroes, Put f(x) = 0
(-1/2) (x + 3)^{2} + 8 = 0
(-1/2)(x + 3)^{2} = -8
Multiplying by -2 on both sides, we get
(x + 3)^{2} = -8 (-2)
(x + 3)^{2} = 16
x + 3 = √16
x + 3 = ±4
Decomposing into two branches, we get
x + 3 = 4 x = 4 - 3 x = 1 |
x + 3 = -4 x = -4 - 3 x = -7 |
So, the zeroes are -7 and 1.
Problem 5 :
f(x) = x^{2} - 24x + 144
Solution :
To find zeroes, Put f(x) = 0
x^{2} - 24x + 144 = 0
x^{2} - 12x - 12x + 144 = 0
x(x - 12) - 12(x - 12) = 0
(x - 12) (x - 12) = 0
Equating each factor to 0, we get
x = 12.
Problem 6 :
f(x) = 9x^{2} - 81
Solution :
To find zeroes, Put f(x) = 0
9x^{2} - 81 = 0
9x^{2} = 81
Dividing by 9 on both sides.
x^{2} = 9
x = √9
x = ±3
So, the zeroes are -3 and 3.
Problem 7 :
f(x) = x^{2} - x - 20
Solution :
To find zeroes, Put f(x) = 0
x^{2} - x - 20 = 0
x^{2} - 5x + 4x - 20 = 0
By grouping
x(x - 5) + 4(x - 5) = 0
(x + 4) (x - 5) = 0
Equating each factor to zero, we get
x + 4 = 0 and x - 5 = 0
x = -4 and x = 5
So, the zeroes are -4 and 5.
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