Question

Find the value of k for the following quadratic equation,so that it should have equal roots :kx2+6x+1=0.

Answer 1

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Answer 2

Solution:

kx^2 + 6x + 1 = 0


The factor theorem states that
when f(c) = 0, then x - c is a factor of the polynomial f(x). This also means that c is a root of f(x)

In this case f(x) = kx^2 + 6x + 1

If f(0) = x^2 + 6x + 1

= (0)^2 + 6(0) + 1

= 1

0 is not the root of f(x)

If f(1/2) = x^2 + 6x + 1

= (1/2)^2 + 6(1/2) + 1

= 1/4 + 3 + 1 = 4 1/4

4 1/4 is not the root of f(x)

If f(1/3) = x^2 + 6x + 1

= (1/3)^2 + 6(1/3) + 1

1/9 + 2 + 1 = 3 1/9

3 1/9 is not the root of f(x)

If f(-1/3) = x^2 + 6x + 1

= (-1/3)^2 + 6(-1/3) + 1

= 1/9 - 2 + 1 = 0

(-1/3) is the root of the f(x)

In this case c = (-1/3)

Therefore f(-1/3) = 0

kx^2 + 6x + 1 = 0

⇒ k(-1/3)^2 + 6(-1/3) + 1

⇒ k(-1/3)^2 = - 6(-1/3) - 1

⇒ 1/9k = 2 - 1

⇒ k = 9

Therefore,

9x^2 + 6x + 1 = (3x + 1) (3x + 1)


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