Question : If $\alpha, \beta$ are the roots of $6 x^2+13 x+7=0$, then the equation whose roots are $\alpha^2, \beta^2$ is:
Option 1: $36 x^2-87 x+49=0$
Option 2: $36 x^2-85 x+49=0$
Option 3: $36 x^2-85 x-49=0$
Option 4: $36 x^2+87 x-49=0$
Correct Answer: $36 x^2-85 x+49=0$
Solution :
Given: $6x^2 + 13x + 7 = 0$
⇒ $x^2 + \frac{13x}{6} + \frac{7}{6} = 0$
Now, $α + β = \frac{- 13}{6}$ and $αβ = \frac{7}{6}$
So, $(α + β)^2 = α^2 + β^2 + 2αβ$
⇒ $\frac{169}{36} = α^2 + β^2 + \frac{7}{3}$
⇒ $\frac{169}{36} - \frac{7}{3} = α^2 + β^2$
⇒ $\frac{169 - 84}{36} = α^2 + β^2$
⇒ $\frac{85}{36} = α^2 + β^2$
Also, $x^2 - (α^2 + β^2)x + α^2β^2= 0$
⇒ $x^2 - \frac{85}{36}x + \frac{49}{36} = 0$
⇒ $36x^2 - 85x + 49 = 0$
Hence, the correct answer is $36x^2 - 85x + 49 = 0$.
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