For what value of 'k' will the following quadratic equation: (k+1)x^2– 4kx^2+9=0 have real and equal roots?Solve the equation.
Answers
Given Equation,
(k+1)x^2– 4kx+9=0 is a quadratic equation.
If it has real, and equal roots then,
Discriminant = 0
Comparing the given equation to standard equation ax² + bx + c = 0 gives,
a = k + 1
b = - 4k
c = 9
⇒Discriminant = 0
⇒ b² - 4ac = 0
⇒(-4k)² - 4(k+1)(9) = 0
⇒16k² - 36k - 36 = 0
⇒ 4k² - 9k - 9 = 0
⇒ 4k² - 12k + 3k - 9 = 0
⇒ 4k ( k - 3) + 3 ( k - 3) = 0
⇒ (4k + 3) ( k - 3)=0
⇒ k = - 3/4 or 3.
The equal roots of the equation are -b/2a
If k = 3,
a = k + 1 = 4
b = - 4k = - 12
The roots of the equation would thus be = - b/2a
= 12/2(4)
= 3/2
If k = - 3/4
a = k + 1 = 1/4
b = - 4k = 3
The roots of the equation would thus be = - b/2a
= - 3 / 2(1/4)
= - 3/ 1/2
= - 6.
Answer:
Step-by-step explanation:
Given Equation,
(k+1)x^2– 4kx+9=0 is a quadratic equation.
If it has real, and equal roots then,
Discriminant = 0
Comparing the given equation to standard equation ax² + bx + c = 0 gives,
a = k + 1
b = - 4k
c = 9
⇒Discriminant = 0
⇒ b² - 4ac = 0
⇒(-4k)² - 4(k+1)(9) = 0
⇒16k² - 36k - 36 = 0
⇒ 4k² - 9k - 9 = 0
⇒ 4k² - 12k + 3k - 9 = 0
⇒ 4k ( k - 3) + 3 ( k - 3) = 0
⇒ (4k + 3) ( k - 3)=0
⇒ k = - 3/4 or 3.
The equal roots of the equation are -b/2a
If k = 3,
a = k + 1 = 4
b = - 4k = - 12
The roots of the equation would thus be = - b/2a
= 12/2(4)
= 3/2
If k = - 3/4
a = k + 1 = 1/4
b = - 4k = 3
The roots of the equation would thus be = - b/2a
= - 3 / 2(1/4)
= - 3/ 1/2
= - 6.