Find the range of values of $k$ for which the quadratic equation $9x^2+8x-2k=0$ has 2 distinct real roots.

Question

How do I do I do this question? I've wrote down $b^2-4ac>0$ because it has 2 distinct real roots. And put the numbers into that equation $8^2-4(9)(-2k)$ and got $64+72k=0$. What do I need to do next please to work out the values of $k$ (answer the question)?

Answer 1

The solutions to $9x^2 +8x-2k = 0$ will have solutions $x = \frac {-8 +\sqrt {8^2 -4\times 9\times (-2k)}}{2\times 9}$ and $x = \frac {-8 -\sqrt {8^2 -4\times 9\times (-2k)}}{2\times 9}$

Sometimes those two answers will be two distinct answers, sometimes those answers will only be one single answer, sometimes those answers will never exist and all.

so the question is: What must be true about $k$ for those to be two different answers and not both the same or not existing at all.

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Okay, those answers will but distinct when $-\sqrt{8^2 - 4\times 9\times (-2k)} \ne +\sqrt{8^2 -4\times 9\times (-2k)}$. Those answer will be the same answer if $-\sqrt{8^2 - 4\times 9\times (-2k)} = +\sqrt{8^2 -4\times 9\times (-2k)}$. And those answers wont exist if $-\sqrt{8^2 - 4\times 9\times (-2k)}$ or $ +\sqrt{8^2 -4\times 9\times (-2k)}$ don't exist.

So when do those happen?

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Okay, those answers will not exist and all of $\sqrt {8^2 -4\times 9 \times (-2k)}$ is the square root of a negative number. They will be equal if $-\sqrt{8^2 - 4\times 9\times (-2k)} = +\sqrt{8^2 -4\times 9\times (-2k)}$ which will happen if $\pm \sqrt{8^2 - 4\times 9\times (-2k)} = 0$. And they'll be different values if it is the square root of a positive numbers.

So: There will be two distinct solutions if $8^2 -4\times 9 \times (-2k)$ is positive.

There will be one double root if $8^2 -4\times 9 \times (-2k)=0$.

And there will be no solutions if $8^2 -4\times 9 \times (-2k)$ is negative.

So you need to solve: $8^2 -4\times 9 \times (-2k) > 0$

$64 + 72k > 0$

$72k > -64$

$k > -\frac {64}{72}=-\frac 89$.

If $k > -\frac 89$ then there are two distinct solutions. If $k = -\frac 89$ there is one double root. If $k< - \frac 89$ there are no real solutions (and two complex solutions).

If $k > -\frac 89$ then that two solutions are $x=\frac {-8\pm\sqrt{64+72k}}{18}= \frac {-8\pm 2\sqrt{16+18k}}{18}=\frac {-4\pm\sqrt{2(8+9k)}}9$. If $k = -\frac 89$ then the two solutions are $x =x=\frac {-8\pm\sqrt{64+72(-\frac 89)}}{18} = \frac {-8\pm\sqrt{64-8\times 8}}{18}=\frac {-8\pm 0}{18}=-\frac 49$. And if $k > \frac 89$ then $x=\frac {-8\pm\sqrt{64+72k}}{18}=\frac {-8\pm\sqrt{(-1)(72|k| -64)}}{18}= \frac {-8\pm 2\sqrt{18|k|-16}i}{18}=\frac {-4\pm \sqrt{2(9|k|-8)}i}2$; two non real roots (where $i$ is the purely imaginary number so that $i^2 = -1$.)