Solve a quadratic for a unique solution.

Question

If $ax^2-5x+2=0$ then find the value of $a$ that makes the quadratic have a unique root.

Answer 1

Hint: Write $$a=\frac{5x+2}{x^2}$$

Answer 2

If $a=0$ it‘s a linear equation and therefore has exactly one solution: $x=\frac{2}{5}$

If $a\neq 0$ you can solve it as a quadratic equation:

$$x = \frac{5\pm \sqrt{25-8a}}{2a}$$

this has exactly one solution iff

$$\sqrt{25-8a} = 0 \Leftrightarrow 25-8a = 0 \Leftrightarrow a=\frac{25}{8}$$

Now the only solution is

$$x = \frac{5}{2*\frac{25}{8}}=\frac{4}{5}$$