Choose the correct answer from the given four options :
If the equation 3x² – kx + 2k =0 roots, then the the value(s) of k is (are)
3x² – kx + 2k = 0Here, a = 3, b = -k, c = 2k
b2 – 4ac
= (–k)2 – 4 x 3 x 2k
= k2 – 24k∴ Roots are equal.
∴ b2 – 4ac = 0
∴ k2 – 24k = 0⇒ k(k – 24) = 0Either k = 0ork - 24 = 0,then k = 24
∴ k = 0, 24.
Concept: Nature of Roots of a Quadratic Equation
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Page 2
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If the equation {k + 1)x² – 2(k – 1)x + 1 = 0 has equal roots, then the values of k are
(k + 1)x² – 2(k – 1)x + 1 = 0Here, a = k + 1, b = -2(k – 1), c = 1
∴ b2 – 4ac
= [–2(k –- 1)]2 – 4(k + 1)(1)
= 4(k2 – 2k + 1) – 4k - 4
= 4k2 – 8k + 4 – 4k – 4
= 4k2 – 12k∵ Roots are equal.
∴ b2 – 4ac = 0
∴ 4k2 – 12k = 04k(k – 3) = 0⇒ 4k(k – 3) = 0⇒ k(k – 3) = 0Either k = 0ork – 3 = 0,then k = 3
k = 0, 3.
Concept: Nature of Roots of a Quadratic Equation
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Page 3
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If the equation 2x² – 6x + p = 0 has real and different roots, then the values of p are given by
p < `(9)/(2)`
p ≤ `(9)/(2)`
p > `(9)/(2)`
p ≥ `(9)/(2)`
2x² – 6x + p = 0Here, a = 2, b = –6, c = p
b2 – 4ac
= (–6)2 – 4 x 2 x p= 36 – 8p∵ Roots are real and unequal.
∴ b2 – 4ac > 0
⇒ 36 – 8p > 0⇒ 36 > 8p
⇒ `(36)/(8)` > p
⇒ p < `(36)/(8)`
⇒ p < `(9)/(2)`.
Concept: Nature of Roots of a Quadratic Equation
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Page 4
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The quadratic equation `2x^2 - sqrt(5)x + 1` = 0 has
two distinct real roots
two equal real roots
no real roots
more than two real roots
`2x^2 - sqrt(5)x + 1` = 0 Here, a = 2, b = `-sqrt(5)`, c = 1
b2 - 4ac = `(-sqrt(5))^2 - 4 xx 2 x 1`
= 5 - 8 = -3∵ b2 - 4ac < 0
∴ It has no real roots.
Concept: Nature of Roots of a Quadratic Equation
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