Question

If one zero of the polynomial p(x)=x^(2)+a is -5 .find the value of a and the other zero. (Hint: Take p(-5)=0 ​

Answer 1

GIVEN :–

• One zero of polynomial p(x) = x² + a is -5.

TO FIND :–

• Other zero of quadratic polynomial.

SOLUTION :–

• If one zero of polynomial is -5 then x = -5 will satisfy the equation.

$\\ \implies \sf p (- 5) = 0$

$\\ \implies \sf {( - 5)}^{2} + a= 0$

$\\ \implies \sf 25 + a= 0$

$\\ \implies \large{ \boxed{ \sf a= - 25}}$

• So that quadratic polynomial is –

$\\ \implies \large{ \boxed{ \sf p (x) = {x}^{2} - 25}}$

• Now Let's find roots –

$\\ \implies \sf {x}^{2} - 25 = 0$

$\\ \implies \sf {x}^{2} - {5}^{2} = 0$

$\\ \implies \sf ( {x}- {5})(x + 5) = 0 \: \: \: \: \: \: \: [ \: \because \: {a}^{2} - {b}^{2} = (a + b)(a - b) ]$

$\\ \implies \sf x = - 5 \: , \: 5$

• Hence , other root of polynomial is 5 .

Answer 2

Given:

• Zero of polynomial : p(x) = x² + a is -5

To find:

• a
• other zero of the polynomial

Solution:

Given ,

Hint : Take p( - 5 )

As per the question given that root is -5.Now, substituting x = - 5 in p(x) and finding the value of a

→ p( -5) = ( - 5)² + a

→ 0 = 25 + a

→ a = -25

Now,

→ p(x) = x² + a

Putting p(x) = 0 & substituting the value of a

→ (x)² - 25 = 0

→ x² = 25

→ x = √25

→ x = 5

Hence, other root of x = 5 & a = -25