Q)  Find the value of m for which the quadratic equation (m + 1) y² – 6(m+1) y + 3(m+9) = 0, m ≠ -1 has equal roots. Hence find the roots of the equation.               

[CBSE, Sample Question Paper – Standard – 2024]

Ans: Given quadratic equation is: (m + 1) y² – 6(m+1) y + 3(m+9) = 0

If we compare to ax² +bx + c = 0, then we can see that

a = (m+1),

b = – 6 (m+1)

and c = 3 (m+9)

For a quadratic equation to have real (positive) and equal roots, the discriminant D should be $0$.

or b² – 4ac = 0

Substituting the values of a, b and c, we get:

∴ (- 6(m+1))² – 4(m+1) x 3(m+9) = 0

∴ 36(m+1)² – 12(m+1)(m+9) = 0

∴ 3(m²+ 2m + 1) – (m² +10m + 9) = 0

∴ 3m²+ 6m + 3 – m² -10m – 9 = 0

∴ 2m² -4m – 6 = 0

∴ m² – 2m – 3 = 0

∴ (m – 3)(m + 1) = 0

m = 3, m = -1

Since m ≠ -1, Hence, m = 3.

Therefore, for m = 3, the given quadratic equation will have equal roots

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Check:  Let’s put the value m = 3 in the given quadratic equation and check if we get 2 equal roots:

(m + 1) y² – 6(m+1) y + 3(m+9) = 0

∴ (3+1)y²– 6 (3 +1)y + 3(3 + 9) = 0

∴ 4 y²– 24 y + 36 = 0

∴ y²– 6 y + 9 = 0

∴ (y -3)²  = 0

It gives two equal roots of y = 3. Hence, our answer m = 3 is correct.

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