FIND ZEROES OF QUADRATIC POLYNOMIAL

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

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

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.

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

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.