Question

find the zeros of the polynomial x square - 5 and verify the relationship between the zeros and the coefficient​

Answer 1

Given polynomial is p(x)=x² - 5

By Remainder Theorem,

p(x)=0

→x² - 5 =0

→x² - (√5)² = 0

→(x+√5)(x - √5)=0

→x=√5 or - √5

Let @ and ß be the zeros of p(x)

→@=√5 and ß= -√5

Now,

Sum of Roots :

@ + ß = √5+(-√5)=0

Product of Zeros:

@ß = (√5)(- √5) = -5

Hence, Verified

Answer 2

Answer :

Verified.

Explanation :

Given :

⇒ p(x) = x² - 5.

To Do :

Verify the relationship between the zeros and the coefficient​.

Solution :

This type of questions can be solved by the 'Remainder Theorem.'

$\bigstar$ Remainder Theorem :

When a polynomial, f(x), is divided by a linear polynomial , x - a, the remainder of that division will be equivalent to f(a).

We can write it as -

⇒ p(x)=0

Now,

$\sf \\\implies x^2 - 5 =0\\ \\\implies x^2 - (\sqrt5)^2 = 0\\ \\\implies (x+\sqrt5)(x - \sqrt5)=0\\ \\\implies x=\sqrt5 \;or\; - \sqrt5$

★ Consider the -

The zeros of p(x)  as - α and ß

Where, α = √5 & ß= -√5.

Now,

Sum of Roots :  

α + ß

=> √5+(-√5)=0

Product of Zeros :

α × ß = (√5)(- √5) = -5  

Hence,

Verified