Question

If the roots of $$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$$ are equal, then, a, b, c are in
A) A.P.
B) G.P.
C) H.P.
D) None of these

Answer 1

Hint: At first, we will find the discriminant of the given quadratic equation.
The nature of the roots of a quadratic equation depends on the discriminant.
From the discriminant, we can find the relationship between $$a,b,c$$.
Also, we know that, if $$p,q,r$$ are in G.P., then, $$q^2=pr$$

Complete step-by-step answer:
It is given that; the roots of $$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$$ are equal.
We have to find out the relation between $$a,b,c$$.
Let us consider a quadratic equation $$a{x}^2+bx+c=0,a\ne 0$$. So, the discriminant of this equation is, $$b^2-4ac$$.
When the roots of a quadratic equation are equal, then the discriminant is zero.
So, we have, $$b^2-4ac=0$$
Since, the roots of $$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$$ are equal, then its discriminant is also be zero.
Substitute the values in the discriminant we get,
$\Rightarrow$$$4b^2(a+c{)}^2=4(a^2+b^2)(b^2+c^2)$$
Simplifying we get,
$\Rightarrow$$$b^2(a^2+c^2+2ac)=a^2{b}^2+b^4+a^2{c}^2+b^2{c}^2$$
Simplifying again we get,
$\Rightarrow$$$a^2{b}^2+b^2{c}^2+2a{b}^{2}c=a^2{b}^2+b^4+a^2{c}^2+b^2{c}^2$$
Rearranging the terms which is equal to 0, we get,
$\Rightarrow$$$b^4-2a{b}^{2}c+a^2{c}^2=0$$
Simplifying again we get,
$\Rightarrow$$$(b^2-2ac{)}^2=0$$
Simplifying we get,
$\Rightarrow$$$b^2=ac$$
We know that, if $$p,q,r$$ are in G.P., then, $$q^2=pr$$
So, we can say that, $$a,b,c$$ are in G.P.

Hence, the correct option is B.

Note: Quadratics or quadratic equations can be defined as a polynomial equation of a second degree. The general form of a quadratic equation is $$a{x}^2+bx+c=0,a\ne 0$$.
The nature of the roots of a quadratic equation depends on the discriminant.
The discriminant is $$D=b^2-4ac$$.
If, $$D>0$$, then the roots of the quadratic equation are real and different.
If, $$D<0$$, then the roots of the quadratic equation are imaginary.
If, $$D=0$$, then the roots of the quadratic equation are real and equal.