# FIND ZEROES OF THE QUADRATIC POLYNOMIAL

Every quadratic polynomial will have two zeroes, those two values only will make the given quadratic polynomial as zero.

Other names of zeroes :

x-intercepts, solutions, values of x, roots.

Problem 1 :

Find the zeroes of the polynomial

4x² - 3x - 1

Solution :

By solving this quadratic equation, we will get zeroes.

4x² - 3x - 1 = 0

4x² - 4x + x - 1 = 0

4x(x - 1) + 1(x - 1) = 0

(4x + 1) (x - 1) = 0

4x + 1 = 0   x - 1 = 0

x = -1/4 and x = 1

Let p(x) = 4x² - 3x - 1

p(x) (4x + 1) (x - 1)

p(-1/4) = (4(-1/4) + 1) (-1/4 - 1)

p(-1/4) = 0

p(1) = (4(1) + 1) (1 - 1)

p(1) = 0

Problem 1 :

Which are the zeros of p(x) = x² + 3x - 10.

a)  5, -2   b)  -5, 2  c)  -5, -2  d)   none of these

Solution :

Let p(x) = 0

x² + 3x - 10 = 0

x² + 5x - 2x - 10 = 0

x(x + 5) - 2(x + 5) = 0

(x + 5) (x - 2) = 0

x + 5 = 0   x - 2 = 0

x = -5        x = 2

Hence, zeros of p(x) is -5, 2.

So, option (b) is correct.

Problem 2 :

Which are the zeros of p(x) = 6x² - 7x + 12.

a)  5, -2   b)  -5, 2  c)  -5, -2  d)  none of these

Solution :

Let p(x) = 0

6x² - 7x + 12 = 0

The expression does not factor.

So, option (d) is correct.

Problem 3 :

Which are the zeros of p(x) = x² + 7x + 12.

a)  4, -3   b)  -4, 3  c)  -4, -3  d)  none of these

Solution :

Let p(x) = 0

x² + 7x + 12 = 0

x² + 4x + 3x + 12 = 0

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

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

x + 4 = 0   x + 3 = 0

x = -4        x = -3

Hence, zeros of p(x) is -4, -3.

So, option (c) is correct.

Problem 4 :

If the product of zeros of the polynomial ax² - 6x - 6 is 4, find the value of ‘a’.

Solution :

ax² - 6x - 6

Product of zeros = c/a

Here a = a, b = -6, c = -6

Product of zeros = -6/a

-6/a = 4

a = -3/2

Hence, value of a is -3/2.

Problem 5 :

If one zero of the polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other. Find the value of a.

Solution :

(a² + 9)x² + 13x + 6a

Product of zeros = c/a

Here a = a² + 9 and c = 6a

α × 1/α = 6a/a² + 9

1 = 6a/a² + 9

a² + 9 = 6a

a² - 6a + 9 = 0

(a - 3)² = 0

a - 3 = 0

a = 3

Hence, the value of a is 3.

Problem 6 :

Find the zeros of the quadratic polynomial x² + 5x + 6 and verify the relationship between the zeros and coefficients.

Solution :

= x² + 5x + 6

= x² + 3x + 2x + 6

= x(x + 3) + 2(x + 3)

= (x + 2) (x + 3)

x + 2 = 0   x + 3 = 0

x = -2          x = -3

So, α = -2 and β = -3

Sum of zeros = α + β = -2 - 3 = -5

Product of zeros = αβ = (-2)(-3) = -6

Problem 7 :

Find the zeros of the polynomial

p(x) = √2x² - 3x - 2√2.

Solution :

p(x) = √2x² - 3x - 2√2

Let p(x) = 0

√2x² - 3x - 2√2 = 0

-2√2(√2) = -4

1(-4) = -4,1 - 4 = -3

(√2x + 1) (√2x - 4) = 0

√2x = -1 and x = -1/√2

√2x - 4 = 0 and x = 4/√2

x = 2√2              x = -1/√2

Hence, x = 2√2 and x = -1/√2.

Problem 8 :

If α, β are the zeros of the polynomials

f(x) = x² + 5x + 8

then α + β

a)  5  b)  -5  c)  8  d)  none of these

Solution :

f(x) = x² + 5x + 8

Here a = 1, b = 5 and c = 8

Sum of zeros (α + β) = -b/a

= -5/1

α + β = -5

So, option (b) is correct.

Problem 9 :

If α, β are the zeros of the polynomials

f(x) = x² + 5x + 8, then α ∙ β

a)  0  b)  1  c)  -1  d)  none of these

Solution :

f(x) = x² + 5x + 8

Here a = 1, b = 5 and c = 8

Product of zeros αβ = c/a

= 8/1

αβ = 8

So, option (d) is correct.

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