A quadratic equation with an unknown parameter

Question

I have a simple formula$$ 3x^2+kx+7=0.$$

I know that the discriminant of the function where the value of $k$ needs to be $ -2{\sqrt21} < k $

I want to find values of $k$ in which the function has two real roots or one real root.

I used the formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$

to eventually get $${\frac{-k \pm \sqrt{k^2-4(3)(7)} }{6}} = 0.$$

I don't know how to get $k$ on its own. I keep getting stuck.

Help much appreciated.

Hint: you will get complex roots if $b^2-4ac < 0$. – Andy Walls Feb 16 '18 at 00:46 — Andy Walls, Feb 16 '18 at 00:46
Hint: Solve $b^2-4ac\ge 0$. – farruhota Feb 16 '18 at 00:48 — farruhota, Feb 16 '18 at 00:48

score 1 · Accepted Answer · answered Feb 16 '18 at 00:47

In order for the equation to have two solutions, you need the discriminant to be positive. This means that : $D>0$. For the case of one solution (double), you need $D=0$. For no solutions at all (over the real numbers), you need $D<0$. Take into account that you only have to work with the expression of $D$ which is the quantity under the root, so you don't have to work around $k$ from the solutions formula.

Specifically, for your question, you need to solve :

$$D>0 \Rightarrow k^2 - 84 >0$$

$$D=0 \Rightarrow k^2-84 = 0$$

for the cases of two real roots and one (double) real root respectively.

A quadratic equation with an unknown parameter

1 Answers1