Question

Find the value of k, if the roots of (k-3)x^2+6x+9=0 are equal ​

Answer 1

Answer:

k = 4.

Explanation:

If roots are equal, b² - 4ac = 0.

6² - 4(k - 3)(9) = 0

36 - 36k + 108 = 0

144 - 36k = 0

-36k = -144

k = 4.

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Answer 2

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Explanation:

by comparing the equation (k-3)x^2+6x+9=0 or x^2+6x/(k-3)+9/(k-3)=0 with quadratic equation x^2-(a+b)x+ab ( where a & b are root of the equation ) we get ,

-(a+b) = 6/(k-3) .. (1) & ab = 9/(k-3) . ... (2)

now, -(a+b) = 6/(k-3)

a+b = -6/(k-3)

a+a = -6/(k-3) [ because roots a and b are equal ]

2a = -6/(k-3)

a = -6/2(k-3) = -3/(k-3)

by putting the value of a in equation (2)

ab = 9/(k-3) . ... (2) or a×a = 9/(k-3) . ... (2) or a^2 = 9/(k-3) . ... (2)

[-3/(k-3)]^2 = 9/(k-3)

9/(k-3)^2 = 9/(k-3)

1/(k-3) = 1

k-3 = 1

k = 1+3 = 4

so k = 4