Question

if a and b are the zeros of the polynomial X²+ 6x + 5 then (a+b)²=___​

Answer 1

Question :-

If a and b are the zeroes of the polynomial x² + 6 x + 5  then (a+b)² =  ?  

Solution :-

given quadratic polynomial is

   → x² + 6 x + 5

As we know

$\bf{sum \:of\:zeroes=\frac{-coefficient\:of\:x}{coefficient\:of\:x^2}}$

so,

$\tt{a+b=\frac{-6}{1}=-6}$

we have to find

→   ( a + b )²

→   ( - 6 )²

→ 36  ( Ans.)

Answer 2

AnsWer :

36.

Solution :

We have, Polynomial.

$\tt \dagger \: \: \: \: \: {x}^{2} + 6x + 5.$

Where as,

• a = 1.
• b = 6.
• c = 5.

We know that,

$\tt \blacksquare Sum \: of \: Zeros. \\ \tt \alpha + \beta = \frac{ - b}{a} = \dfrac{Coefficient \: of \: x }{Coefficient \: of \: {x}^{2} }$

$\tt \blacksquare Product \: of \: Zeros. \\ \tt \alpha \beta = \frac{ c}{a} = \dfrac{Constant \: term }{Coefficient \: of \: {x}^{2} }$

$\rule{100}3$

A/Q,

$\tt : \implies{(a + b)}^{2} = {( \alpha + \beta) }^{2}$

$\tt : \implies {(a + b)}^{2} = {[\dfrac{- b}{a} ] }^{2}$

$\tt : \implies {(a + b)}^{2} = {[\dfrac{ - 6}{1} ]}^{2}$

$\tt : \implies {(a + b)}^{2} = 36.$

Therefore, the required answer is 36.