Question

please help me to find answers ............Find the zeros of the polynomial given below and verify the relationship between the zeros and its coefficients .a) x×x-5 b) 8x×x-4​

Answer 1

Question

Find the zeros of the polynomial given below and verify the relationship between the zeros and its coefficients .a) x² - 5 b) 8x² - 4.

Solution

Let us assume that the zeros of the polynomial be a and b.

a) x² - 5

(x² + 0x - 5 = 0

Here, a = 1, b = 0 and c = -5)

→ x² - 5 = 0

→ x² = 5

→ x = ±√5

Therefore, the zeros are a = +√5 and b = -√5

Verification

Sum of zeros = -b/a

√5 + (-√5) = 0/1

√5 - √5 = 0

0 = 0

Product of zeros = c/a

(√5) × (-√5) = -5/1

-5 = -5

b) 8x² - 4

(8x² + 0x - 4 = 0

Here, a = 8, b = 0 and c = -4)

→ 8x² - 4 = 0

→ 4(2x² - 1) = 0

→ 2x² - 1 = 0

→ 2x² = 1

→ x² = 1/2

→ x = ± √(1/2)

→ x = ± 1/√2

Therefore, (zeros) a = +1/√2 and b = -1/√2

Verification

Sum of zeros = -b/a

+1/√2 + (-1/√2) = 0/8

+1/√2 - 1/√2 = 0

0 = 0

Product of zeros = c/a

(1/√2) × (-1/√2) = -4/8

-1/2 = -1/2

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

$\leadsto\mathsf \red {(a)\:x^{2} - 5 }$

Given polynomial is:

• $\mathsf {{x}^{2} - 5}$

Let the zeros be α,β. (α>β)

$\mathsf {{x }^{2} - 5 = 0}$

$\leadsto \mathsf {{x}^{2} = 5}$

$\leadsto \mathsf {x = \pm \sqrt{5} }$

$\alpha = \sqrt{5}$

$\beta = - \sqrt{5}$

So,

• α+β = 0
• αβ = -5

The coefficient of x is zero:

• α + β = 0

Constant = αβ = -5

Hence verified.

$\leadsto\mathsf \blue {(b)\:8x^{2} - 4}$

We know the identity a² - b² = (a+b)(a-b)

• a² - b² = (a+b)(a-b)

Using it , we can write : 8x² - 4

$\implies \mathsf {4(2x^2 -1)}$

$\implies \mathsf{4[(\sqrt{2x})^2 - 1^2]}$

$\implies \mathsf {4(\sqrt{2x} + 1)\:(\sqrt{2x} - 1)}$

So, the value of 8x² - 4 is zero

When x = -1/√2 and 1/√2

Verification:

Compare 8 x² - 4 with ax²+ bx + c,

$\underline \mathsf \red {We\:Get\: :-}$

• a = 8
• b = 0
• c = -4

(i) Sum of the zeroes

$\implies$ -1/√2+1/√2

$\implies$ 0

$\implies$ -(Coefficient of x)/(Coefficient of x²)

(ii) Product of the zeroes = (-1/√2)×(1/√2)

$\implies \mathsf {\frac{ -1}{2} }$

$\implies$(Constant term)/(Coefficient of x²)