Question

X²+7x+10 verify the relationship ​

Answer 1

${\tt{\pink{\underline{\large{Given}}}}}$

A polynomial X²+7x+10

${\sf{\blue{\underline{\large{To\:Verify}}}}}$

Relationship between the cofficient

${\sf{\orange{\underline{\Large{Explanation}}}}}$

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

Then,

$\tt\alpha{+}\beta\frac{-b}{a}$

&

$\tt\alpha{\times}\beta{=}\frac{c}{a}$

Here,

a=1

b=7

C=10

$\tt\alpha{+}\beta=\dfrac{{-b}}{a}\dfrac{-1}{7}$

$\tt\alpha{+}\beta{=}\dfrac{-(Cofficient\:of\:X)}{Cofficient\:of\:x^2}$

&

$\tt\alpha{\times}\beta{=}\dfrac{10}{1}$

$\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

→Hence,verified✓✓

Answer 2

Given :

• A polynomial x²+7x+10.

To Find :

• Zeroes of the polynomial and verify its relationship.

Solution :

$\longmapsto\tt\bf{{x}^{2}+7x+10}$

By Splitting Middle Term :

$\longmapsto\tt{{x}^{2}+5x+2x+10}$

$\longmapsto\tt{x(x+5)+2(x+5)}$

$\longmapsto\tt{(x+2)\:(x+5)}$

• x = -2
• x = -5

So , -2 and -5 are the zeroes of polynomial x²+7x+15...

_______________________

Here :

• a = 1
• b = 7
• c = 10

Sum of Zeroes :

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{-2+(-5)=\dfrac{-7}{1}}$

$\longmapsto\tt{-2-5=-7}$

$\longmapsto\tt\bf{-7=-7}$

Product of Zeroes :

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{-2\times{-5}=\dfrac{10}{1}}$

$\longmapsto\tt\bf{10=10}$

HENCE VERIFIED