# FIND ZEROES OF QUADRATIC POLYNOMIAL

The quadratic equation be in any of the forms given below.

• Standard form
• Vertex form
• Factored form

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.

(i) Factoring

(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 = 5x = 5 + 4x = 9 x - 4 = -5x = -5 + 4x = -1

So, the zeroes are 9 and -1.

Problem 2 :

y = 2x2 - 9x - 5

Solution :

y = 2x2 - 9x - 5

To find zeroes, put y = 0

0 = 2x2 - 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 = 02x = -1x = -1/2 x - 5 = 0x = 5

So, the zeroes are -1/2 and 5.

Problem 3 :

f(x) = 3x2 + 5x - 12

Solution :

To find zeroes, Put f(x) = 0

3x2 + 5x - 12 = 0

3x2 + 9x - 4x - 12 = 0

3x(x + 3) - 4(x + 3) = 0

(3x - 4)(x + 3) = 0

Equating each factor to 0.

 3x - 4 = 03x = 4x = 4/3 x + 3 = 0x = -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 = 4x = 4 - 3x = 1 x + 3 = -4x = -4 - 3x = -7

So, the zeroes are -7 and 1.

Problem 5 :

f(x) = x2 - 24x + 144

Solution :

To find zeroes, Put f(x) = 0

x2 - 24x + 144 = 0

x2 - 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) = 9x2 - 81

Solution :

To find zeroes, Put f(x) = 0

9x2 - 81 = 0

9x2 = 81

Dividing by 9 on both sides.

x2 = 9

x = √9

x = ±3

So, the zeroes are -3 and 3.

Problem 7 :

f(x) = x2 - x - 20

Solution :

To find zeroes, Put f(x) = 0

x2 - x - 20 = 0

x2 - 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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