Question

Find the zero of the polynomial t^2 -15 and verify the relationship between the zero and coefficient

Answer 1

AnswEr:

$\star\;\bold{\underline{\sf{Given\: Polynomial\: : \: t^2 \:-15}}}$

$\implies\sf t^2 - (\sqrt{15})^2$

$\:\:\:\:\:\:\:\small\bold{\underline{\sf{\pink{Using\: Identity\; a^2 - b^2 \:=\: (a \:-\:b) (a\:+\:b)}}}}$

$\implies\sf (t - \sqrt{15}) (t + \sqrt{15}) = 0$

$\implies\sf t = \sqrt{15} \:\&\; t = - \sqrt{15}$

$\sf{\therefore\; Zeroes\; of \; the \: Given \; polynomial \; are \; \sqrt{15} \:\&\; -\sqrt{15}}$

$\implies\sf \alpha = \sqrt{15} \:\& \: \beta = - \sqrt{15}$

$\rule{150}2$

$\:\;\:\:\;\;\;\;\;\;\;\dag\bold{Sum\: of \; Zeroes}$

$\implies\sf \alpha + \beta = -\dfrac{b}{a}$

$\implies\sf \sqrt{15} - \sqrt{15} = -\dfrac{0}{1}$

$\implies\sf 0 = 0$

$\:\;\:\:\;\;\;\;\;\;\;\dag\bold{Product\: of \; Zeroes}$

$\implies\sf \alpha \;\beta = \dfrac{c}{a}$

$\implies\sf (\sqrt{15}) (-\sqrt{15}) = \dfrac{-15}{1}$

$\implies\sf -(\sqrt{15})^2 = \dfrac{-15}{1}$

$\implies\sf -15 = -15$

$\large\bold{\underline{\sf{\purple{Hence,\: verified!}}}}$

Answer 2

Explanation :

$\rule{100}2$

Concept :

So basically, in this question we've to find zeroes of the polynomial t² - 15.

But something is missing here, isn't it?

We know that, 15 can be written as ( √15 )² . Now move towards it's solution :-

Solution :

$\sf t^2 - 15$

$\sf = t^2 - (\sqrt{15})^2$

$\sf = ( t - \sqrt{15}) (t+\sqrt{15})$

For finding zeroes :

t - √15 = 0

⇒ t = √15

∴ α = √15

Also,

t + √15 = 0

⇒ t = - √15

∴ β = - √15

Finally,

Verification :

Here, a = 1; b = 0 ; c = - 15 __________ ( from equation )

We know that,

sum of zeroes ( α + β ) = - b /a

⇒ √15 + ( - √15 ) = 0/1

⇒ 0 = 0

L.H.S = R.H.S

Also,

Product of zeroes ( αβ ) = c/a

⇒ √15 × ( - √15 ) = - 15/1

⇒ -15 = -15

L.H.S = R.H.S

Hence verified!

$\rule{200}2$