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CAT 2024 Slot 2 — QA Question 4

Number TheoryEasy

If x and y are real numbers such that 4x2 + 4y2 - 4xy - 6y + 3 = 0, then the value of (4x + 5y) is

Answer & solution

Answer: 7

Solution

Easy

A single equation in two real variables that has a definite numeric answer is almost always a hidden sum of squares. Group the terms so the whole expression is forced to zero.

1

Write as a sum of squares. Group 4x2−4xy+y24x^2-4xy+y^2 as (2x−y)2(2x-y)^2 and the rest as a multiple of (y−1)2(y-1)^2.

4x2+4y2−4xy−6y+3⇒ (2x−y)2+3y2−6y+3(took out (2x−y)2)⇒ (2x−y)2+3(y−1)2=0\begin{aligned} &4x^2+4y^2-4xy-6y+3\\ &\Rightarrow\ (2x-y)^2+3y^2-6y+3 \quad\text{(took out }(2x-y)^2\text{)}\\ &\Rightarrow\ (2x-y)^2+3(y-1)^2=0 \end{aligned}
2

Each square must vanish. A sum of two non-negative squares is zero only if both are zero.

y−1=0 ⇒ y=12x−y=0 ⇒ 2x=1 ⇒ x=12(using y=1)\begin{aligned} &y-1=0\ \Rightarrow\ y=1\\ &2x-y=0\ \Rightarrow\ 2x=1\ \Rightarrow\ x=\tfrac12 \quad\text{(using }y=1\text{)} \end{aligned}
3

Compute the required value.

4x+5y=4(12)+5(1)=2+5=7\begin{aligned} &4x+5y=4\left(\tfrac12\right)+5(1)=2+5=7 \end{aligned}
4x+5y=74x+5y=7
CAT 2024 Slot 2 QA Q4: If x and y are real numbers such that 4x 2 + 4y 2 - 4xy - 6y + 3 = 0, then the value of (4x + 5y) is — Solution | TheCATExam