Question

Find the value of 'a" if one zero of polynomial (a^2 +1)x^2 + 56x + 2a is reciprocal of the other.​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

•One zero of the quadratic polynomial

(a² + 1)x² + 56x + 2a is reciprocal to other.

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

•The value of " a "

$\;\;\underline{\textbf{\textsf{ Formula used :-}}}$

•Product of zeroes = $\sf \dfrac{constant \: term}{coefficient \: of \: x^2}$

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Here, we know

•Product of zeroes = $\sf \dfrac{constant \: term}{coefficient \: of \: x^2}$

Let the zeroes of the quadratic polynomial be α and 1/α.

$\longrightarrow \sf \alpha \times \frac{1}{\alpha} = \frac{2a}{a^2 +1}\\\\ \longrightarrow \sf 1 = \frac{2a}{a^2 +1}\\\\\longrightarrow \sf a^2 +1=2a \\\\\longrightarrow \sf a^2 - 2a + 1 = 0$

$\:\:\:\:\dag\bf{\underline \green{Using\:splitting\:middle \:term:-}}$

( It’s is a method for factoring quadratic equations.)

$\sf \longrightarrow a^2 - a - a + 1 = 0\\\\\\\sf \longrightarrow a(a-1)-1(a-1)= 0 \\\\\longrightarrow \sf (a-1)(a-1)= 0\\\\\longrightarrow \sf a = 1$

$\:\:\:\:\dag\bf{\underline{\underline \green{Hence:-}}}$

The value of a is = 1

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