JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let $α, β$ be the roots of the equation $x^{2} - 3 x + r = 0$ , and $\frac{α}{2}, 2 β$ be the roots of the equation $x^{2} + 3 x + r = 0$ . If the roots of the equation $x^{2} + 6 x = m$ are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2}$ , then $m$ is equal to :-

A
$- 135$
B
$- 567$
C
$135$
D
$567$

Solution

From the first equation the sum of roots gives $α + β = 3$ . From the second equation the sum of its roots gives $\frac{α}{2} + 2 β = - 3$ .

Solving these two linear equations simultaneously yields $α = 6$ and $β = - 3$ .

The product of roots of the first equation is $α β = r$ , so $r = - 18$ .

Rewriting the third equation as $x^{2} + 6 x - m = 0$ , the product of its roots equals $- m$ . Substituting the values:

$2 α + β + 2 r = 12 - 3 - 36 = - 27$ and $α - 2 β - \frac{r}{2} = 6 + 6 + 9 = 21$ .

Therefore $- m = (- 27) (21) = - 567$ , giving $m = 567$ .

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