Reciprocated Roots

Algebra Level 2

There is a quadratic equation with roots that are reciprocals of each other. If the leading coefficient is 2, find the value of the constant.

FYI, this was in this year's NTSE (don't know which state).

The answer is 2.

2 solutions

Shubhendra Singh
Nov 3, 2014

Let the polynomial be = $ax^{2} + bx +c$ where $a=2$

Since the roots are resiprocal of each other

Product of roots =1 = $\frac{c}{a}$

By this $c=2$

BTW, this was not in my paper. So, rather than NTSE, you must specify which state paper. Like HTSE for Haryana etc.

Kartik Sharma - 6 years, 7 months ago

What was your result?

Sharky Kesa - 6 years, 7 months ago

Abdur Rehman Zahid
Nov 24, 2014

The solutions of the equation are of the form $\color{#D61F06}{a,\frac{1}{a}}$ ,so the equation is $\color{#3D99F6}{(x-a)(x-\frac{1}{a})=x^2-(\frac{1}{a}+a)x+1}$ .But the leading coefficient is $2$ so we have to multiply the equation by $2$ .So the equation is $\color{#D61F06}{2x^2-2(\frac{1}{a}+a)x+2}$ .So the constant term is $\boxed{2}$

Reciprocated Roots

FYI, this was in this year's NTSE (don't know which state).

The answer is 2.

2 solutions

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