Quadratic Functions and Equations

SAT Questions that focus on Quadratic Functions and Equations require knowledge of the following topics.

A quadratic function is any function that can be rearranged in the form: f(x) = ax² + bx + c, where "a" is a real number, a ≠ 0. "a", "b", and "c" are called coefficients.

Examples:
f(x) = 3x² + 7x + 2
f(x) = -2x² + 4
y = x² + 3x
y = -8x²

Solutions of the quadratic equation

Considere the quadratic equation $ax^2+bx+c=0$

Multiplying both sides by "4a":

$4a^2x^2+4abx+4ac=0$

Now let's add b² to both sides:

$(4a^2x^2+4abx+b^2)+4ac=b^2$
$(4a^2x^2+4abx+b^2)=b^2-4ac$
$(2ax+b)^2=b^2-4ac$
$2ax+b=^+_-\sqrt{b^2-4ac}$

$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$

This is called the Bhaskara Formula.

The expression $b^2-4ac$ is called the discriminant of the quadratic equation, and is usually represented by the greek letter $\Delta$.

If $\Delta$>0, the equation has two disstinct real solutions (or roots);
if $\Delta$=0, the equation has only one real solution (or root);
if $\Delta$<0, the equation has no real solution (or root).

Graph of the quadratic equation

The graph of a quadratic equation is a parabola.

If coefficient a>0, the parabola opens to the top.
If coefficient a<0, the parabola opens to the bottom.

If $\Delta$>0, the parabola crosses the x-axis (the equation has two distinct real roots).
If $\Delta$=0, the parabola touches (but does not cross) the x-axis (the equation has only one real root).
If $\Delta$<0, the parabola does not touch the x-axis (the equation has no real root).

Consider the equation $x^2+4x+2$.

Coefficient "a" of this equation is 1 (positive), thus the parabola opens to the top.
Coefficient "b" is 4 and coefficient "c" is 2. Therefore its $\Delta$ is 4²-4*1*2 =8. The parabola should cross the x-axis (the equation has two distinct real solutions).

The graph of this equation is:




Consider now the equation $-x^2+2x-2$.

Coefficient "a" is -1 (negative), thus the parabola opens to the bottom.
Coefficient "b" is 2 and coefficient "c" is -2. Therefore its $\Delta$ is 2²-4*(-1)*(-2) =-4. Since the $\Delta$ of the equation is negative, its graph does not cross the x-axis (the equation has no real solution).

The graph of this equation is:

The vertex of the Parabola

The graph of the quadratic equation is mirror-symmetrical. Therefore its vertex is in the midpoint between the two roots of the equation.

The roots of the quadratic equation are give by the formula:

$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$

The mean of the two roots is:

$x_v=\frac{\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}}{2}$

$x_v=\frac{\frac{-b}{2a}+\frac{-b}{2a}}{2}$

$x_v=\frac{\frac{-2b}{2a}}{2}$

$x_v=-\frac{b}{2a}$

Other ways to solve quadratic equations

Factoring by simple inspection
In some cases it is possible to express the quadratic equation as (px+q)(rx+s)=0, where p, q, r, and s are real numbers. The solutions to the quadratic equation are the solutions to px+q=0 and to rx+s=0. For example, $x^2+3x+2=0$ can be written as (x+1)(x+2)=0; the solutions are x=-1 and x=-2.

Completing the square
In some cases it is easier to express the quadratic equation as $(x+r)^2=p$, where p and r are real numbers. For example,
$2x^2-8x-2=0$. Dividing both sides by 2:
$x^2-4x+-1=0$. Adding 5 to both sides:
$x^2-4x-1+5=5$
$x^2-4x+4=5$
$(x-2)^2=5$
$x=2+\sqrt5$ or $x=2-\sqrt5$

Solved SAT Practice Tests

Find Practice Tests in the following links:

SAT Practice Tests - Quadratic Functions and Equations

Additional Practice Tests - Quadratic Functions and Equations