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प्रश्न
Find the value of k so that sum of the roots of the quadratic equation is equal to the product of the roots:
(k + 1)x2 + (2k + 1)x - 9 = 0, k + 1 ≠ 0.
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उत्तर
The given equation is
(k + 1)x2 + (2k + 1)x - 9 = 0
Here, a = k + 1, b = (2k + 1) and c = -9.
Sum of the roots α + β = `(- (2k + 1))/(k + 1)`
and αβ = `c/a = (-9)/(k + 1)`
Since, Sum of the roots = Product of the roots
Then, `((2k +1)/(k + 1)) = (9)/(k + 1)`
⇒ 2k + 1 = 9
⇒ 2k = 9 - 1
⇒ 2k = 8
⇒ k = `(8)/(2)`
= 4
⇒ k = 4.
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Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
Δ = 4 + `square` = 40
∴ b2 – 4ac > 0
∴ The roots of the equation are real and unequal.
