Theory of EquationshardPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

EE Main 2026 Quadratic Roots Problem: Find m = 567 | Solution

JEE Maths question with a full step-by-step solution.

Question
Let α,β\alpha,\beta be the roots of the equation x2−3x+r=0x^2-3x+r=0, and α2, 2β\dfrac{\alpha}{2},\ 2\beta be the roots of the equation x2+3x+r=0x^2+3x+r=0. If the roots of the equation x2+6x=mx^2+6x=m are 2α+β+2r2\alpha+\beta+2r and α−2β−r2\alpha-2\beta-\dfrac{r}{2}, then mm is equal to
A−135-135
B−567-567
C135135
D567567correct
Solution
Step 1: From x2−3x+r=0x^2-3x+r=0, sum and product of roots give
α+β=3.\alpha+\beta=3.
 Step 2: From x2+3x+r=0x^2+3x+r=0, the sum of its roots α2\dfrac{\alpha}{2} and 2β2\beta is −3-3:
α2+2β=−3.\frac{\alpha}{2}+2\beta=-3.
 Step 3: Solving the two equations α+β=3\alpha+\beta=3 and α2+2β=−3\dfrac{\alpha}{2}+2\beta=-3 together:
α=6,β=−3.\alpha=6,\qquad \beta=-3.
 Step 4: Product of roots of the first equation equals rr:
αβ=r ⇒ r=(6)(−3)=−18.\alpha\beta=r\ \Rightarrow\ r=(6)(-3)=-18.
 Step 5: Now for x2+6x−m=0x^2+6x-m=0, the product of its two given roots equals −m-m:
−m=(2α+β+2r)(α−2β−r2).-m=(2\alpha+\beta+2r)\left(\alpha-2\beta-\frac{r}{2}\right).
 Step 6: Substitute α=6, β=−3, r=−18\alpha=6,\ \beta=-3,\ r=-18:
2α+β+2r=12−3−36=−27,α−2β−r2=6+6+9=21.2\alpha+\beta+2r=12-3-36=-27,\qquad \alpha-2\beta-\frac{r}{2}=6+6+9=21.
 Step 7: Therefore
−m=(−27)(21)=−567 ⇒ m=567.-m=(-27)(21)=-567\ \Rightarrow\ m=567.
 (Check: sum of roots =−27+21=−6=-27+21=-6, matching the coefficient +6+6 in x2+6x−m=0x^2+6x-m=0.) Correct answer: (4)
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