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Maths Question 1 – JEE-MAIN 2026

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Let α,β be the roots of the equation x2−3x+r=0, and α2,2β be the roots of the equation x2+3x+r=0. If the roots of the equation x2+6x=m are 2α+β+2r and α−2β−r2, then m is equal to:

HINT 1: Vieta's Formulas. The relationship between the coefficients of a polynomial and the sums and products of its roots is the key to solving this problem.

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Ninja Strategy•Sign Analysis

By determining the signs of the roots of the final equation without full calculation, we can determine the sign of m and eliminate the negative options.

● Video Walkthrough
Step 1: Find the values of α, β, and r✦ Active

For the equation x2−3x+r=0 with roots α,β, Vieta's formulas give us:

α+β=3(1) αβ=r(2)

For the equation x2+3x+r=0 with roots α2,2β, the sum of roots is:

α2+2β=−3⟹α+4β=−6(3)

Subtracting equation (1) from (3), we get 3β=−9, so β=−3. Substituting this back into (1), we find α−3=3, so α=6. Finally, from (2), r=αβ=(6)(−3)=−18.

Step 2: Determine the roots of the third equation○ Expand

The third equation is x2+6x=m, which can be written as x2+6x−m=0. Its roots are R1=2α+β+2r and R2=α−2β−r2. We substitute the values α=6,β=−3,r=−18:

R1=2(6)+(−3)+2(−18)=12−3−36=−27
R2=6−2(−3)−−182=6+6+9=21
Step 3: Calculate the value of m○ Expand

For the quadratic equation x2+6x−m=0, the product of the roots R1 and R2 is equal to the constant term, −m.

R1⋅R2=(−27)(21)=−567

Therefore, −m=−567, which implies m=567.

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