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प्रश्न
Solve (x2 + 3x)2 - (x2 + 3x) -6 = 0.
उत्तर
(x2 + 3x)2 - (x2 + 3x) -6 = 0
Putting x2 + 3x = y, the given equation becomes
y2 - y - 6 = 0
⇒ y2 - 3y + 2y - 6 = 0
⇒ y(y - 3) + 2(y - 3) = 0
⇒ (y - 3) (y + 2) = 0
⇒ y - 3 = 0 or y + 2 = 0
⇒ y = 3 or y = -2
But x2 + 3x = y
x2 + 3x = 3
⇒ x2 + 3x - 3 = 0
Here a = 1, b = 3, c = -3
Then x = `(-b ± sqrt(b^2 - 4ac))/(2a)`
x = `(-3 ± sqrt(9 + 12))/(2)`
x = `(-3 ± sqrt(21))/(2)`
or
x2 + 3x = -2
x2 + 3x + 2 = 0
x2 + 2x + x + 2 = 0
x (x + 2) +1(x + 2) = 0
x + 2 = 0 or x + 1 = 0
x = -2 or x = -1
Hence, roots are `(-3 ± sqrt(21))/(2), -2, -1`.
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