Find the value of $a$ from the equation

Question

The roots of the equation: $ax^2-(5a+2)x+9a=0$ are equal.

Find the value of $a$ given that $a>0$.

Answer 1

It's a normal quadratic equation in $x$. How do solve those? Is there some point in solving those where you have a test that tells you how many solutions it has? (The roots being equal is equivalent to it having only one solution).

Answer 2

Hint #$1$:

$$\frac{(5a+2)\color{red}{+}\sqrt{(5a+2)^2-4a\cdot9a}}{2a}=\frac{(5a+2)\color{red}{-}\sqrt{(5a+2)^2-4a\cdot9a}}{2a}$$

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Hint #$2$:

$$\color{red}{+}\sqrt{n}=\color{red}{-}\sqrt{n}\implies{n}=0$$

Answer 3

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is the quantity $\Delta = b^2 - 4ac$. The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by the Quadratic Formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a}$$ Observe that the two roots are equal if and only if $\Delta = 0$.

For the quadratic equation $ax^2 - (5a + 2)x + 9a = 0$, with $a > 0$, the discriminant is $$\Delta = [-(5a + 2)]^2 - 4a(9a)$$ Set the discriminant equal to zero, solve for $a$, and select the positive solution for $a$.