# Thread: [SOLVED] Show the roots of this equation : how?

1. ## [SOLVED] Show the roots of this equation : how?

Show that the roots of this quadratic equation $(k + 5)x^2 + 13x -6k = 0$ are real and distinct for all the real values of k.

This was how I did it.

a = (K + 5), b = 13, c = -6k

Using b² - 4ac,

$(13)² - 4(k + 5)(-6k)$
$= 169 + 24k² + 120k$
$= 24k^2 + 120k + 169$

Using the formula of finding the roots of a quadratic equation,

$\frac{-120 \pm \sqrt{(120)^2 - 4(24)(169)}}{2(24)}$

$\frac{-120 \pm \sqrt{-1824}}{2(24)}$

After this, I don't know what to do. I know that this way is wrong so please correct me on the steps.

2. Originally Posted by mark1950
Show that the roots of this quadratic equation $(k + 5)x^2 + 13x -6k = 0$ are real and distinct for all the real root values of k.

...
Solve this equation for x using the quadratic formula:

$x=\dfrac{-13 \pm \sqrt{169-4(k+5)\cdot(-6k)}}{2(k+5)}$

The roots are real if the radical is positive (or zero) for all values of k:

$x=\dfrac{-13 \pm \sqrt{169+24k^2+120k}}{2(k+5)}$

$x=\dfrac{-13 \pm \sqrt{24(k^2+5k + \dfrac{25}{4})-\dfrac{24\cdot25}{4}+169}}{2(k+5)}$

$x=\dfrac{-13 \pm \sqrt{24(k^2+\dfrac52)^2+19}}{2(k+5)}$

Thus the radical is at least +19 and therefore there must exist real roots of the equation.

3. Originally Posted by mark1950
Show that the roots of this quadratic equation $(k + 5)x^2 + 13x -6k = 0$ are real and distinct for all the real root values of k.
I'm not sure that the bolded word should be there.

This was how I did it.

a = (K + 5), b = 13, c = -6k

Using b² - 4ac,

$(13)² - 4(k + 5)(-6k)$
$= 169 + 24k² + 120k$
$= 24k^2 + 120k + 169$

Using the formula of finding the roots of a quadratic equation,

$\frac{-120 \pm \sqrt{(120)^2 - 4(24)(169)}}{2(24)}$
If the roots of $(k + 5)x^2 + 13x -6k = 0$ are real and distinct, then it should be $24k^2 + 120k + 169 {\color{red}>} 0$, not $24k^2 + 120k + 169 {\color{red}=} 0$.

When solving a quadratic inequality, the first thing I do is change it to an equation anyway to find the critical points. So assuming that you plugged in numbers into the quadratic formula right:
$\frac{-120 \pm \sqrt{-1824}}{2(24)}$

This tells you that the quadratic $24k^2 + 120k + 169 = 0$ has no real roots, that it doesn't cross the x-axis. That means that for any value of k, the expression $24k^2 + 120k + 169$ will always be positive or always be negative. Since the leading coefficient is positive, $24k^2 + 120k + 169 > 0$ for any k. Therefore, the roots of $(k + 5)x^2 + 13x -6k = 0$ are real and distinct.

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4. oh.......I get it now. Lol. Thanks, man!

5. Originally Posted by mark1950
For all values of k, 24k² + 120k + 169 > 0. This, I now know. But how does this nature of k influence the roots of the quadratic equation (k + 5)x² + 13x - 6k = 0 to be real and distinct? Examples would be helpful. Thanks.
The discriminant of the quadratic equation
$(k + 5)x^2 + 13x - 6k = 0$,
as you have shown, is
$24k^2 + 120k + 169$.

Recall that for the discriminant b² - 4ac:
if b² - 4ac > 0, there are two distinct real roots,
if b² - 4ac = 0, there is one real double root, and
if b² - 4ac < 0, there are no real roots.

Since we have shown that $24k^2 + 120k + 169 > 0$ for any real number k, there will be two distinct real roots in the quadratic $(k + 5)x^2 + 13x - 6k = 0$, for any real number k.

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