Question

If one zero of the polynomial (a^2+a)x^2+13x+6 a is reciprocal of the Other find the value of a

Answer 1

$\textbf{Answer :}$

The given polynomial is

p (x) = (a² + a) x² + 13x + 6a

Let, the zeroes are α and 1/α

Then,

α + 1/α = - 13/(a² + a) ...(i)

and

α × 1/α = 6a/(a² + a)

⇒ 1 = 6/(a + 1)

⇒ a + 1 = 6

⇒ a = 6 - 1

⇒ a = 5

∴ the value of a is 5

#$\textbf{MarkAsBrainliest}$

Answer 2

Heya !!!



Let one zero or the given polynomial Be Alpha


Other zero will be ( 1/Alpha )




P(X) = (A²+A)X² + 13X² + 6A



Here,


A = ( A²+A) , B = 13 and C = 6A





Product of zeroes = C/A



Alpha × 1/ Alpha = 6A/ ( A²+A)




1 = 6A/ ( A²+A)




A² + A = 6A



A² = 6A - A



A² = 5A


A × A = 5A



A = 5A/A




A = 5.





HOPE IT WILL HELP YOU....... :-)