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CAT 2023 Slot 2 — QA Question 2

IndicesEasy

The sum of all possible values of x satisfying the equation 24x2 - 22x2+x+16 + 22x+30 = 0, is

Answer & solution

Correct answer: 1/2

  • A

    3

  • B

    5/2

  • C

    3/2

  • 1/2

Solution

Easy

The three terms are a perfect-square trinomial in disguise: A2−2AB+B2=(A−B)2A^2-2AB+B^2=(A-B)^2. Spot AA and BB, collapse to a single square, set its base to zero, and solve the resulting quadratic in xx.

1

Rewrite each term with a common eye on 22x22^{2x^2} and 2x+152^{x+15}:

24x2=(22x2)222x2+x+16=2⋅22x2⋅2x+1522x+30=(2x+15)2\begin{aligned} &2^{4x^2}=\left(2^{2x^2}\right)^2\\ &2^{2x^2+x+16}=2\cdot 2^{2x^2}\cdot 2^{x+15}\\ &2^{2x+30}=\left(2^{x+15}\right)^2 \end{aligned}
2

Recognise the perfect square with A=22x2A=2^{2x^2}, B=2x+15B=2^{x+15}:

A2−2AB+B2=0⇒ (22x2−2x+15)2=0⇒ 22x2=2x+15\begin{aligned} &A^2-2AB+B^2=0\\ &\Rightarrow\ \left(2^{2x^2}-2^{x+15}\right)^2=0\\ &\Rightarrow\ 2^{2x^2}=2^{x+15} \end{aligned}
3

Equate exponents and solve the quadratic:

2x2=x+15⇒ 2x2−x−15=0⇒ (2x+5)(x−3)=0⇒ x=−52  or  x=3\begin{aligned} &2x^2=x+15\\ &\Rightarrow\ 2x^2-x-15=0\\ &\Rightarrow\ (2x+5)(x-3)=0\\ &\Rightarrow\ x=-\tfrac52\ \text{ or }\ x=3 \end{aligned}
4

Add the roots (or use sum =−ba=12=-\frac{b}{a}=\frac12 directly):

3+(−52)=123+\left(-\tfrac52\right)=\tfrac12
Sum of values of x=12— option (d).\text{Sum of values of }x=\mathbf{\tfrac12}\quad\text{— option (d).}

Once you reach 2x2−x−15=02x^2-x-15=0, skip factoring: the sum of roots is −ba=−−12=12-\dfrac{b}{a}=-\dfrac{-1}{2}=\dfrac12.

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CAT 2023 Slot 2 QA Q2 — Indices solution | TheCATExam