Advertisements
Advertisements
प्रश्न
Choose the correct answer from the given four options :
If the equation {k + 1)x² – 2(k – 1)x + 1 = 0 has equal roots, then the values of k are
पर्याय
1, 3
0, 3
0, 1
0, 1
Advertisements
उत्तर
(k + 1)x² – 2(k – 1)x + 1 = 0
Here, 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 = 0
4k(k – 3) = 0
⇒ 4k(k – 3) = 0
⇒ k(k – 3) = 0
Either k = 0
or
k – 3 = 0,
then k = 3
k = 0, 3.
APPEARS IN
संबंधित प्रश्न
Find the values of k for the following quadratic equation, so that they have two equal roots.
2x2 + kx + 3 = 0
Find the values of k for which the roots are real and equal in each of the following equation:
5x2 - 4x + 2 + k(4x2 - 2x - 1) = 0
For what value of k, (4 - k)x2 + (2k + 4)x + (8k + 1) = 0, is a perfect square.
If the roots of the equation (a2 + b2)x2 − 2 (ac + bd)x + (c2 + d2) = 0 are equal, prove that `a/b=c/d`.
If the roots of the equations ax2 + 2bx + c = 0 and `bx^2-2sqrt(ac)x+b = 0` are simultaneously real, then prove that b2 = ac.
Solve the following quadratic equation using formula method only
`5/4 "x"^2 - 2 sqrt 5 "x" + 4 = 0`
Determine, if 3 is a root of the given equation
`sqrt(x^2 - 4x + 3) + sqrt(x^2 - 9) = sqrt(4x^2 - 14x + 16)`.
Find the value of k for which the given equation has real roots:
kx2 - 6x - 2 = 0
Given that one root of the quadratic equation ax2 + bx + c = 0 is three times the other, show that 3b2 – 16ac.
If b = 0, c < 0, is it true that the roots of x2 + bx + c = 0 are numerically equal and opposite in sign? Justify.
