Author Topic: QQ >>>> quick quadratic q (Read 1742 times) Tweet Share

lacoste · « **on:** February 15, 2009, 04:42:40 pm »

ahow that the equation (k+1)x^2-2x-k=0 has a solution for all values of k.

thanks~!!!

cobby · « **Reply #1 on:** February 15, 2009, 04:49:31 pm »

Is the equation supposed to read
$(k+1)^2-2x-k=0$

or

$(k+1)x^2 -2x-k =0$

/0 · « **Reply #2 on:** February 15, 2009, 04:52:31 pm »

By 'a' solution I'm assuming 1 or more solutions.

The discriminant $\Delta =b^2 - 4ac$ can be used to find the number of solutions. If $b^2 - 4ac \geq 0$ , then the equation will have solutions.

$\Delta = (-2)^2 - 4(k+1)(-k)$

$= 4 + 4k(k+1)$

$= 4k^2+4k+4$

$= 4(k^2+k+1)$

$= 4(k^2 + k + \frac{1}{4} - \frac{1}{4} +1)$

$=4\left((k+\frac{1}{2})^2 + \frac{3}{4}\right)$

$= 4(k+\frac{1}{2})^2 + 3$

Since $(k+\frac{1}{2})^2\geq 0$ , we can see that $4(k+\frac{1}{2})^2 +3 \geq 3$ . So the equation will always have 2 solutios for any value of k.

lacoste · « **Reply #3 on:** February 15, 2009, 06:51:51 pm »

thanks divide 0!!

cobby: the second one the x^2

Quote from: cobby on February 15, 2009, 04:49:31 pm

Is the equation supposed to read
$(k+1)^2-2x-k=0$

or

$(k+1)x^2 -2x-k =0$

lacoste · « **Reply #4 on:** February 15, 2009, 07:30:15 pm »

hey divide0, did you accidently place 3 instead of +3/4 at the end of the working? or did something cancel out which i dont know where?

d0minicz · « **Reply #5 on:** February 15, 2009, 10:24:41 pm »

Before he completed the square, he took 4 out of the equation as a factor.
When hes completing the square, the 4 still remains outside as a factor.
so the $\frac {3}{4}$ cancels down to 3.

lacoste · « **Reply #6 on:** February 15, 2009, 10:56:02 pm »

thanks dominicz.

get it now. 12/4=3 cheers

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Author Topic: QQ >>>> quick quadratic q (Read 1742 times) Tweet Share

lacoste

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