Use the quadratic formula and discriminant

For

the quadratic formula is

The SAT reference sheet doesn't include this formula, so you need to know it by heart. Four spots in it are easy to get wrong:

• Start with $-b$, the opposite of $b$. If $b$ is negative, $-b$ is positive.
• Keep the $\pm$. When the square root is positive, it gives two answers, one with $+$ and one with $-$.
• Put all of $b^2-4ac$ under the square root, including the $-4ac$.
• Put $2a$ under the whole numerator. It divides $-b$ too, not only the square root.

Before you substitute, make one side equal $0$. Then write $a$, $b$, and $c$ on their own line, signs included. It feels like an extra step, but it's where sign slips get caught. For

that line is

Substitute with each value in parentheses, so every sign stays put:

Watch the last part under the root. $-4(5)(-3)$ has two negatives, so it's positive: $+60$. Now finish:

The $\pm$ keeps both exact roots: $\frac{7+\sqrt{109}}{10}$ and $\frac{7-\sqrt{109}}{10}$. The square root stays because $109$ isn't a perfect square, so these roots are irrational. A perfect square like $49$ would have simplified to $7$, leaving no square root.

The discriminant is the part under the square root:

To count real solutions, you only need its sign. Think about what it does to the $\pm$:

• If $D>0$, $\sqrt D$ is a positive number. Adding it and subtracting it give two different real solutions.
• If $D=0$, $\sqrt D=0$. Adding $0$ and subtracting $0$ give the same number, so there's one real solution.
• If $D<0$, $\sqrt D$ isn't a real number, since no real number squared is negative. So there are no real solutions.

An easy way to remember it: positive, zero, negative means two, one, none.

With integer coefficients, a positive $D$ that is a perfect square gives rational roots. Any other positive $D$ leaves a square root, so the roots are irrational.

On a graph, the same three cases are cross, touch, miss: the parabola crosses the $x$-axis twice, touches it once, or misses it. The middle case is easy to misread. The parabola doesn't cross the axis, but it does touch it at one point, where $y=0$. So that point counts as one real solution. It's called a repeated root, because the $+$ and $-$ versions of the formula both land on it.

A graph shows these cases well. But when the answer is the exact constant that makes a parabola just touch the axis, don't estimate it with a slider, the Desmos control you drag to change a constant. Set $D=0$ and solve, like you did for $m$ at the start.

The quadratic formula works on every quadratic. But completing the square can be shorter when the $x^2$ term has a coefficient of $1$ and the $x$ coefficient is easy to cut in half. The idea is to add the right number so one side becomes a perfect square, like $(x+4)^2$. It can feel like a trick the first time, but it's the same four moves every time: move the constant, halve the $x$ coefficient and square it, add that to both sides, and take the square root.

Take

Move the constant to the right:

Half of $8$ is $4$, and $4^2=16$. Add $16$ to both sides:

Why $16$? Because

so $16$ is exactly the piece the left side was missing.

Now take the square root of both sides, keeping both signs:

The two solutions are $-4+\sqrt{21}$ and $-4-\sqrt{21}$.

If the $x^2$ coefficient isn't $1$, divide every term by it first, not only the $x^2$ term. This works best when the division leaves simple numbers.