Find the value of k for which the quadratic equation kx(x-3)+9=0 has equal and real roots.
Answers
Answer:
k = 4
Step-by-step explanation:
Given, Quadratic Equation is kx(x - 3) + 9 = 0.
⇒ kx² - 3kx + 9 = 0
On comparing with ax² + bx + c = 0, we get a = k, b = -3k, c = 9.
Given that the equation has real equal and real roots.
∴ D = 0
b² - 4ac = 0
⇒ (-3k)² - 4(k)(9) = 0
⇒ 9k² - 36k = 0
⇒ 9k² = 36k
⇒ 9k = 36
⇒ k = 36/9
⇒ k = 4
Hope it helps!
Given :
The quadratic equation is : k x ( x - 3 ) + 9 = 0
==> k x² - 3 k x + 9 = 0
We need to find the value of "k" such that the equation has real and equal roots .
Comparing k x² - 3 k x + 9 = 0 with a x² + b x + c = 0 : -
==> a = k............................(1)
==> b = - 3 k..........................(2)
==> c = 9..........................(3)
For an equation to have real and equal roots :
b² must be equal to 4 a c
We assume the value of k such that this equation has real and equal roots :
Thus : [ see ( 1 ) , ( 2 ) and ( 3 ) ] b² = 4 a c
==> ( - 3 k )² = 4 × k × 9
==> 9 k² = 9 × 4 k
==> k² = 4 k [ cancelling 9 both sides ]
==> k = 4 [cancelling k both sides ]
The value of k was found to be 4.
Hope it helps you :-)
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