Advertisements
Advertisements
Question
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.
Advertisements
Solution
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.
RELATED QUESTIONS
Without solving, examine the nature of roots of the equation x2 – 5x – 2 = 0
Determine the nature of the roots of the following quadratic equation:
`3x^2-2sqrt6x+2=0`
Find the values of k for which the roots are real and equal in each of the following equation:
(2k + 1)x2 + 2(k + 3)x + (k + 5) = 0
In the following determine the set of values of k for which the given quadratic equation has real roots:
2x2 + kx - 4 = 0
Solve the following quadratic equation using formula method only :
16x2 = 24x + 1
Solve the following quadratic equation using formula method only
`3"x"^2 +2 sqrt 5 "x" -5 = 0`
Find the value of k for which the roots of the equation 3x2 -10x +k = 0 are reciprocal of each other.
The sum of the roots of the quadratic equation 3x2 – 9x + 5 = 0 is:
Every quadratic equation has exactly one root.
One root of equation 3x2 – mx + 4 = 0 is 1, the value of m is ______.
