Advertisements
Advertisements
प्रश्न
For what values of k, the roots of the quadratic equation (k + 4) x2 + (k + 1) x + 1 = 0 are equal ?
उत्तर
The given quadratic equation is (k + 4)x2 + (k + 1)x + 1 = 0.
For equal roots, its discriminant, D is 0.
⇒ b2 − 4ac = 0 where a = k + 4, b = k + 1, c = 1
⇒ (k + 1)2 − 4(k + 4) × 1 = 0
⇒ k2 + 2k + 1− 4k − 16 = 0
⇒ k2 − 2k − 15 = 0
⇒ k2 − 5k + 3k − 15 = 0
⇒ k(k − 5) + 3(k − 5) = 0
⇒ (k − 5) (k + 3) = 0
⇒ k = 5 or k = −3
Thus, for k = 5 or k = −3, the given quadratic equation has equal roots.
APPEARS IN
संबंधित प्रश्न
Write the discriminant of the following quadratic equations:
x2 - 2x + k = 0, k ∈ R
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
3a2x2 + 8abx + 4b2 = 0, a ≠ 0
`11-x=2x^2`
`2x^2-2sqrt2x+1=0`
`4sqrt3x^2+5x-2sqrt3=0`
`2sqrt3x^2-5x+sqrt3=0`
`1/x-1/(x-2)=3,x≠0,2`
`4x^2-4a^2x+(a^4-b^4)=0`
`3a^2x^2+8abx+4b^2=0`
`12abx^2-(9a^2-8b^2)x-6ab=0,` `Where a≠0 and b≠0`