Question

Find the zeroes of the polynomial 4s^2-4s+1 and verify the relationship between the zeroes and coefficient

Answer 1

Solution :

Step : 1

First of all, factories the polynomial 4s² - 4s + 1

$\sf = 4s^2 - 4s +1$

$\sf = 4s^2 - ( 2+2)s+1$

$\sf = 4s^2 - 2s - 2s + 1$

$\sf = 2s(2s-1)-1(2s-1)$

$\sf = (2s-1)(2s-1)$

Step : 2

For finding zeroes of the polynomial :-

2s - 1 = 0

⇒ 2s = 1

⇒ s = 1/2

∴ α = 1/2

∴ β = 1/2

Step 3 :

Verification :

Here, a = 4 ; b = - 4 ; c = 1

We know that the sum of the zeroes of the polynomial is equal to - b/a

Substitute the values. We get,

α + β = - b/a

→ 1/2 + 1/2 = - ( - 4 )/4

→ 2/2 = 4/4

→ 1 = 1

∴ L.H.S = R.H.S

Also,

The product of zeroes of polynomial is equal to c/a,i.e

αβ = c/a

Substitute the given values. We get,

1/2 × 1/2 = 1/4

→ 1/4 = 1/4

∴ L.H.S = R.H.S

Hence verified!

$\rule{200}2$

Answer 2

✯✯ QUESTION ✯✯

Find the zeroes of the polynomial ${4s2}^{2}-4s+1$ and verify the relationship between the zeroes and coefficient..

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✰✰ ANSWER ✰✰

$⇝{4s}^{2}-4s+1$

$⇝{4s}^{2}-4s+1=0$

$⇝{4s}^{2}-2s-2s+1=0$

$⇝2s(2s-1)-1(2s-1)=0$

$⇝2s-1=0$ ︴ $⇝2s-1=0$

$⇝2s=1$ ︴ $⇝2s=1$

$⇝s=\dfrac{1}{2}$ ︴ $⇝s=\dfrac{1}{2}$

⇝$\dfrac{1}{2}\:and\:\dfrac{1}{2}$ are the zeroes of the polynomials...

☛$\small{\boxed{\bold{\bold{\purple{\sf{Sum\:of\:Zeroes=\dfrac{Coff.\:of\:x}{Coff.\:of\:x^2}}}}}}}$

$⇝α +β =\dfrac{-b}{a}$

$⇝\dfrac{1}{2}+\dfrac{1}{2}=-\cancel\dfrac{(-4)}{4}$

$⇝\dfrac{1+1}{2}=\dfrac{1}{1}$

$⇝\cancel\dfrac{2}{2}=\dfrac{1}{1}$

$⇝\dfrac{1}{1}=\dfrac{1}{1}$

☛$\textbf{L.H.S = R.H.S}$

☛$\small{\boxed{\bold{\bold{\red{\sf{Product\:of\:Zerroes=\dfrac{Constant\:Term}{Coff.\:of\:x^2}}}}}}}$

$⇝αβ=\dfrac{c}{a}$

$⇝\dfrac{1}{2}×\dfrac{1}{2}=\dfrac{1}{4}$

$⇝\dfrac{1}{4}=\dfrac{1}{4}$

☛$\textbf{L.H.S = R.H.S}$