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

Answer:

The value of a = 1.

Step-by-step explanation:

Consider the :

The one zero of the following polynomial as - α.

The other zero as - $\bf \dfrac{1}{\alpha}$

Now,

$\underline{\underline{\textbf{We know that}}\;:}}}\\ \\ \\\bigstar\;\boxed{\bf The\;product\;of\;zeros = \dfrac{Constant\;term}{Coefficient\; of\; x^2}}$

So,

$\implies\sf \alpha \times \dfrac{1}{\alpha} = \dfrac{2a}{a^2 +1}\\ \\ \\\implies \sf 1 = \dfrac{2a}{a^2 +1}\\ \\ \\\ \implies \sf a^2 +1=2a \\ \\ \\\implies \sf a^2 - 2a + 1 = 0$

_____________[ Quadratic equation.. ]

We solve the quadratic equation by using splitting the middle term :

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

Hence, The value of a = 1.

$\rule{300}{1.5}$

• About Splitting the middle term :

Splitting the middle term is a method for factoring quadratic equations.

In which, x term is the sum of two factors and product equal to last term.

Answer 2

Answer:

$\boxed{\red{\bf a = 1}}$

Step-by-step explanation:

Given that :

• One zero of the quadratic polynomial (a² + 1)x² + 56x + 2a is reciprocal to other.

To find :

• The value of a.

Solution:

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

We know that,

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

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

We got a quadratic equation, solving this equation by splitting the middle term ;

$\sf \implies a^2 - a - a + 1 = 0\\\\\implies \sf a(a-1)-1(a-1)= 0 \\\\\implies \sf (a-1)(a-1)= 0\\\\\implies \sf a = 1 \: or \: a = 1$

Hence, the required value of a is 1.