# Thread: Find Values of k

1. ## Find Values of k

For which values of k will the roots of the equation

x^2 = 2x(3k + 1) - 7(2k + 3) be equal?

x^2 = 6xk + 2 - 14k - 21

x^2 = 6xk - 14k -19

Is this a new start?

2. ## Re: Find Values of k

For which values of k will the roots of the equation

x^2 = 2x(3k + 1) - 7(2k + 3) be equal?

x^2 = 6xk + 2 - 14k - 21

x^2 = 6xk - 14k -19

Is this a new start?

you should get

$x^2 = (2+6k)x -(14k+21)$

3. ## Re: Find Values of k

Hint: A quadratic of the form \displaystyle \begin{align*} a\,x^2 + b\,x + c = 0 \end{align*} has exactly one root (or two equal roots) when the discriminant \displaystyle \begin{align*} b^2 - 4\,a\,c = 0 \end{align*}.

4. ## Re: Find Values of k

Originally Posted by romsek

you should get

$x^2 = (2+6k)x -(14k+21)$
How did you get +21 when -7 × 3 = -21?
How did you get 14k when -7 × 2k = -14k?

5. ## Re: Find Values of k

Originally Posted by Prove It
Hint: A quadratic of the form \displaystyle \begin{align*} a\,x^2 + b\,x + c = 0 \end{align*} has exactly one root (or two equal roots) when the discriminant \displaystyle \begin{align*} b^2 - 4\,a\,c = 0 \end{align*}.
Are you saying to rewrite x^2=(2+6k)x−(14k+21) in the form ax^2 + bx + c and then use the discriminant by setting it to zero?

Yes

7. ## Re: Find Values of k

Originally Posted by Prove It
Yes
Ok. I will solve it tomorrow.

8. ## Re: Find Values of k

$x^2=2x(3k+1)-7(2k+3)$

$x^2=6kx+2x-14k-21$

$x^2=(6k+2)x-(14k+21)$

$x^2-(6k+2)x+(14k+21)=0$

For roots to be equal

$b^2-4ac=0$

$[-(6k+2)]^2-4×1×(14k+21)=0$

$36k^2+4+24k-56k-84=0$

$36k^2-32k-80=0$

$4(9k^2-8k-20)=0$

$9k^2-8k-20=0$

$(k-2)(9k+10)=0$

$k=2,\;-\dfrac{10}{9}$

9. ## Re: Find Values of k

Originally Posted by deesuwalka
$x^2=2x(3k+1)-7(2k+3)$

$x^2=6kx+2x-14k-21$

$x^2=(6k+2)x-(14k+21)$

$x^2-(6k+2)x+(14k+21)=0$

For roots to be equal

$b^2-4ac=0$

$[-(6k+2)]^2-4×1×(14k+21)=0$

$36k^2+4+24k-56k-84=0$

$36k^2-32k-80=0$

$4(9k^2-8k-20)=0$

$9k^2-8k-20=0$

$(k-2)(9k+10)=0$

$k=2,\;-\dfrac{10}{9}$
Very impressive math work.