disicriment of solutions

**johnsy123** · March 7th 2010, 03:30 AM

$6x^2+4kx+(k+3)=0$

i)find one solution
ii) two
iii)none

**e^(i*pi)** · March 7th 2010, 03:38 AM

Originally Posted by johnsy123

$6x^2+4kx+(k+3)=0$

i)find one solution
ii) two
iii)none

Compare your equation to the standard form of a quadratic equation.

Quadratic: $ax^2+bx+c=0$

Yours: $6x^2 + 4kx + (k+3) = 0$

Assuming k is a constant we can match the coefficients:

$a = 6$
$b = 4k$
$c = k+3$

The discriminant is given by: $\Delta = b^2-4ac$

If:

$\Delta > 0$ then there are two, different, real solutions
$\Delta = 0$ then there are two, equal, real solutions
$\Delta < 0$ then there are no real solutions

I will let you figure out the rest of the question

**Prove It** · March 7th 2010, 03:46 AM

Originally Posted by johnsy123

$6x^2+4kx+(k+3)=0$

i)find one solution
ii) two
iii)none

For all of the above, you need to evaluate the discriminant.

$\Delta = b^2 - 4ac$

$= (4k)^2 - 4(6)(k + 3)$

$= 16k^2 - 24k - 72$ .

For one solution, $\Delta = 0$ .

So $16k^2 - 24k - 72 = 0$

$8(2k^2 - 3k - 9) = 0$

$8(2k^2 - 6k + 3k - 9) = 0$

$8[2k(k - 3) + 3(k - 3)] = 0$

$8(k - 3)(2k + 3)= 0$ .

So $k - 3 = 0$ or $2k + 3 = 0$

$k = 3$ or $k = -\frac{3}{2}$ , for one solution.

No solutions: $\Delta < 0$ .

So $16k^2 - 24k - 72 < 0$ .

To evaluate this, you will need to complete the square.

$16\left(k^2 - \frac{3}{2}k - \frac{9}{2}\right) < 0$

$16\left[k^2 - \frac{3}{2}k + \left(-\frac{3}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 - \frac{9}{2}\right] < 0$

$16\left[\left(k - \frac{3}{4}\right)^2 - \frac{9}{16} - \frac{72}{16}\right] < 0$

$16\left[\left(k - \frac{3}{4}\right)^2 - \frac{81}{16}\right] < 0$

$\left(k - \frac{3}{4}\right)^2 - \frac{81}{16} < 0$

$\left(k - \frac{3}{4}\right)^2 < \frac{81}{16}$

$\sqrt{\left(k - \frac{3}{4}\right)^2} < \sqrt{\frac{81}{16}}$

$\left|k - \frac{3}{4}\right| < \frac{9}{4}$

$-\frac{9}{4} < k - \frac{3}{4} < \frac{9}{4}$

$-\frac{3}{2} < k < 3$ .

2 solutions: $\Delta > 0$ .

So $16k^2 - 24k - 72 > 0$ .

To evaluate this, again you will need to complete the square.

$16\left(k^2 - \frac{3}{2}k - \frac{9}{2}\right) > 0$

$16\left[k^2 - \frac{3}{2}k + \left(-\frac{3}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 - \frac{9}{2}\right] > 0$

$16\left[\left(k - \frac{3}{4}\right)^2 - \frac{9}{16} - \frac{72}{16}\right] > 0$

$16\left[\left(k - \frac{3}{4}\right)^2 - \frac{81}{16}\right] > 0$

$\left(k - \frac{3}{4}\right)^2 - \frac{81}{16} > 0$

$\left(k - \frac{3}{4}\right)^2 > \frac{81}{16}$

$\sqrt{\left(k - \frac{3}{4}\right)^2} > \sqrt{\frac{81}{16}}$

$\left|k - \frac{3}{4}\right| > \frac{9}{4}$

$k - \frac{3}{4} < -\frac{9}{4}$ or $k - \frac{3}{4} > \frac{9}{4}$

$k < -\frac{3}{2}$ or $k > 3$ .

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