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: $36 x^2+87 x-49=0$
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
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 $36x^2 - 85x + 49 = 0$.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-2x+4=0$, then what is the equation whose roots are $\frac{\alpha ^{3}}{\beta ^{2}}$ and $\frac{\beta ^{3}}{\alpha ^{2}}?$
Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-x+1=0$, then which equation will have roots $\alpha ^{3}$ and $\beta ^{3}?$
Question : If $\tan (\alpha+\beta)=\sqrt{3}, \tan (\alpha-\beta)=1$ where $(\alpha+\beta)$ and $(\alpha-\beta)$ are acute angles, then what is $\tan$ $(6 \alpha) ?$
Question : If the sum of the roots of a quadratic equation is 1 and the product of the roots is –20, find the quadratic equation.
Question : If $\alpha+\beta=90^{\circ}$ and $\alpha=2 \beta$, then the value of $3 \cos ^2 \alpha-2 \sin ^2 \beta$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile