Question

find one zero of (a^2 +a) x^2 +13x+6a is reciprocal of other then find value of a

Answer 1

$\textbf{Answer :}$

The given polynomial is

(a^2 + a) x^2 + 13x + 6a

Let, the two roots are A and 1/A.

Then,

A + 1/A = -13/(a^2 + a) ...(i)

and

A × 1/A = 6a/(a^2 + a)

or, 1 = 6a/(a^2 + a)

or, a^2 + a = 6a

or, a^2 + a - 5a = 0

or, a^2 - 4a = 0

or, a (a - 4) = 0

or, a - 4 = 0, since a in not zero

or, a = 4

Therefore, the value of a is 4.
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Why not a = 0?

If a = 0 is possible. Then putting the value a = 0, we get

(0^2 + 0) x^2 + 13x + (6 × 0)

= 0 + 13x + 0

= 13x and we are not getting any quadratic polynomial.
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#$\textbf{MarkAsBrainliest}$

Answer 2

Heya !!!!



Let one zero is Alpha .


Other zero will be 1/ Alpha




P(X) = ( A²+A)X² + 13X + 6A


This equation is of the form of AX²+BX + C = 0

Where,



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....... :-)