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CAT 2023 Slot 1 — QA Question 9

Two Quadratic EquationsEasy

Let α and β be two distinct root of the equation 2x2 – 6x + k = 0, such that (α + β) and αβ are the two roots of the equation x2 + px + p = 0. Then the value of 8(k - p)?

Answer & solution

Correct answer: 6

Answer: 6

Solution

Easy

α,β\alpha,\beta are roots of the first quadratic, so use Vieta's to get α+β\alpha+\beta and αβ\alpha\beta in terms of kk. Those two numbers are themselves the roots of the second quadratic — apply Vieta's again to pin down kk and pp, then compute 8(k−p)8(k-p).

1

Vieta's on 2x2−6x+k=02x^2-6x+k=0:

α+β=62=3αβ=k2\begin{aligned} &\alpha+\beta=\frac{6}{2}=3\\ &\alpha\beta=\frac{k}{2} \end{aligned}
2

(α+β)(\alpha+\beta) and αβ\alpha\beta are the roots of x2+px+p=0x^2+px+p=0. So their sum is −p-p and their product is pp:

3+k2=−p(sum)⋯(1)3⋅k2=p(product)⋯(2)\begin{aligned} &3+\tfrac{k}{2}=-p &&\text{(sum)}\quad\cdots(1)\\ &3\cdot\tfrac{k}{2}=p &&\text{(product)}\quad\cdots(2) \end{aligned}
3

Add (1) and (2) so the ±p\pm p cancels:

3+k2+3k2=−p+p=0⇒ 3+2k=0 ⇒ k=−32⇒ p=3k2=−94from (2)\begin{aligned} &3+\tfrac{k}{2}+\tfrac{3k}{2}=-p+p=0\\ &\Rightarrow\ 3+2k=0\ \Rightarrow\ k=-\tfrac{3}{2}\\ &\Rightarrow\ p=\tfrac{3k}{2}=-\tfrac{9}{4}\quad\text{from }(2) \end{aligned}
4

Plug into the target expression:

8(k−p)=8 ⁣(−32+94)=8⋅34=68(k-p)=8\!\left(-\tfrac{3}{2}+\tfrac{9}{4}\right)=8\cdot\tfrac{3}{4}=6
8(k−p)=68(k-p)=\mathbf{6}

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CAT 2023 Slot 1 QA Q9: Let α and β be two distinct root of the equation 2x 2 – 6x + k = 0, such that (α + &bet — Solution | TheCATExam