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प्रश्न
Find the value of 'p' for which the quadratic equation p(x – 4)(x – 2) + (x –1)2 = 0 has real and equal roots.
उत्तर
Given quadratic equation is
p(x – 4)(x – 2) + (x –1)2 = 0
⇒ p(x2 – 4x – 2x + 8) + (x2 + 1 – 2x) = 0
⇒ px2 – 6px + 8p + x2 + 1 – 2x = 0
⇒ x2(p + 1) – 2x(3p + 1) + (8p + 1) = 0
Comparing the above equation with ax2 + bx + c = 0, we get
a = p + 1, b = –2(3p + 1) and c = 8p + 1
For real and equal roots
D = 0 i.e., b2 – 4ac = 0
∴ [–2(3p + 1)]2 – 4(p + 1)(8p + 1) = 0
⇒ 4(3p + 1)2 – 4(8p2 + 9p + 1) = 0
⇒ 4(9p2 + 1 + 6p) – 32p2 – 36p – 4 = 0
⇒ 36p2 + 4 + 24p – 32p2 – 36p – 4 = 0
⇒ 4p2 – 12p = 0
⇒ 4p(p – 3) = 0
⇒ p = 0 or p = 3
Hence, for p = 0 or p = 3, the given quadratic equation has real and equal roots.
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