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प्रश्न
If the roots of the given quadratic equation are real and equal, then find the value of ‘m’.
(m – 12)x2 + 2(m – 12)x + 2 = 0
Find the value of ‘m’ if the quadratic equation (m – 12)x2 + 2(m – 12)x + 2 = 0 has real and equal roots.
उत्तर
(m – 12)x2 + 2(m – 12)x + 2 = 0
Comparing the above equation with ax2 + bx + c = 0, we get
a = m − 12, b = 2(m − 12), c = 2
∆ = b2 − 4ac
= [2(m − 12)]2 − 4 × (m − 12) × 2
= 4(m – 12)2 – 8(m – 12)
= 4(m2 – 24m + 144) – 8m + 96
= 4m2 – 96m + 576 – 8m + 96
= 4m2 – 104m + 672 ...(1)
∴ m2 – 26m + 168 = 0 ...(dividing by 4 on each side)
∴ m2 – 12m – 14m + 168 = 0 ....`[(168= - 14; -12),(- 14 xx -12 = 168),(- 14 - 12 = - 26)]`
∴ m(m – 12) – 14(m – 12)
∴ (m – 12) (m – 14) = 0
∴ m – 12 = 0 or m – 14 = 0
∴ m = 12 or m = 14
But, m = 12 is invalid because on taking m = 12, the coefficient of x2 = m – 12
= 12 – 12
= 0
Therefore the given equation will not be a quadratic equation.
∴ m = 14
∴ The value of m is 14.
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