Question

In the quadratic equation $a{{x}^{2}}+bx+c=0$, a $\ne $0, $\Delta ={{b}^{2}}-4ac$and $\alpha +\beta ,{{\alpha }^{2}}+{{\beta }^{2}},{{\alpha }^{3}}+{{\beta }^{3}}$ are in G.P. where $\alpha ,\beta $ are the root of $a{{x}^{2}}+bx+c=0$ then
A . $\Delta \ne 0$
B. $b\Delta =0$
C. \[c\Delta =0\]
D. \[\Delta =0\]

Answer 1

Hint: In this question, we are given a quadratic equation with their roots and the series which are in GP. We have to find which option satisfies the given question. We know if a,b,c are in GP then ${{b}^{2}}=ac$. By applying that condition and solving the equation, we come up with two values of \[\alpha \]and \[\beta \]. By taking two cases, we are able to find out the option that satisfies the given equation.

Formula used:
If a, b, and c are in G.P. then ${{b}^{2}}=ac$

Complete step by step Solution:
Given quadratic equation is $a{{x}^{2}}+bx+c=0$
And $\alpha +\beta ,{{\alpha }^{2}}+{{\beta }^{2}},{{\alpha }^{3}}+{{\beta }^{3}}$ are in G.P.
We know if a,b,c are in GP, then ${{b}^{2}}=ac$
Here a = $\alpha +\beta $, b = \[{{\alpha }^{2}}+{{\beta }^{2}}\], c = \[{{\alpha }^{3}}+{{\beta }^{3}}\]
Then ${{({{\alpha }^{2}}+{{\beta }^{2}})}^{2}}=(\alpha +\beta )({{\alpha }^{3}}+{{\beta }^{3}})$
Solving the above equation, we get
${{\alpha }^{4}}+{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}={{\alpha }^{4}}+{{\beta }^{4}}+{{\alpha }^{3}}\beta +\alpha {{\beta }^{3}}$
${{\alpha }^{2}}{{\beta }^{2}}-{{\alpha }^{3}}\beta -\alpha {{\beta }^{3}}+{{\alpha }^{2}}{{\beta }^{2}}=0$
Solving further, we get
${{\alpha }^{2}}\beta (\beta -\alpha )-\alpha {{\beta }^{2}}(\beta -\alpha )=0$
Therefore, $({{\alpha }^{2}}\beta -\alpha {{\beta }^{2}})(\beta -\alpha )=0$
Now we take \[\alpha \beta \] from the above equation, and we get
$\alpha \beta (\alpha -\beta )(\beta -\alpha )=0$
Or $\alpha \beta {{(\alpha -\beta )}^{2}}=0$
That is $\alpha =0,\beta =0$ or $\alpha -\beta =0$
Now we take two cases:-
Case 1 :
When $\alpha =0,\beta =0$
Then x = 0 is a solution of given equation.
That is \[\beta \]$a{{(0)}^{2}}+b(0)+c=0$
Hence c = 0
Case 2 :
When $\alpha -\beta =0$
Then $\alpha =\beta $
So, the equation has real roots.
Then $\Delta ={{b}^{2}}-4ac=0$
Therefore, c = 0 or \[\Delta =0\]
Then \[c\Delta =0\]

Therefore, the correct option is (C).

Note: Let us know the rules of discriminant:
A ) If D is positive (D>0), the solutions are real and distinct.
B ) If D is zero (D = 0), the solutions are real and equal.
C ) If D is negative (D < 0), the solutions are distinct and unreal.
Solutions refer to two roots of the equation. The degree Of the question tells us about the number of solutions.