Introduction to Quadratics: Practice

A quadratic equation must contain a squared term $ax^2$ with $a \ne 0$.

$x^2 - 3x + 2 = 0 \quad (a = 1,\ b = -3,\ c = 2)$

The other three have no $x^2$ term, so they are linear equations.

Substitute each candidate into $x^2 - 9$ and look for $0$.

$3^2 - 9 = 9 - 9 = 0$

So $x = 3$ is a solution. (The other root is $-3$, which is not listed.)

By the zero-product property, one of the factors must be $0$.

$x - 4 = 0 \quad \text{or} \quad x - 1 = 0$

So $x = 4$ or $x = 1$.

Set the factor equal to $0$ and solve the linear equation.

$x - 5 = 0 \;\Rightarrow\; x = 5$

The factor $x - 5$ is zero when $x = 5$.

Take the square root of both sides and keep both signs.

$x = \pm\sqrt{16} = \pm 4$

So $x = 4$ or $x = -4$. Writing only $x = 4$ would lose the negative root.

The only number whose square is $0$ is $0$ itself.

$x^2 = 0 \;\Rightarrow\; x = 0$

So there is exactly one solution, $x = 0$, a single repeated root.

Standard form has every term on one side, equal to $0$, with the $x^2$ term written first.

$x^2 + 2x - 8 = 0$

The factored $(x+4)(x-2)=0$ is a product, and the other two are not set equal to $0$.

A product is $0$ when a factor is $0$, so read the roots off the factors.

$x - 2 = 0 \;\Rightarrow\; x = 2$

So $x = 2$ is a solution. (The other root, $-5$, is not among the choices.)

Take the square root of both sides, keeping both signs.

$x - 1 = \pm 3$

Then $x = 1 + 3 = 4$ or $x = 1 - 3 = -2$.

Since $36 > 0$, two different numbers square to $36$.

$x = \pm 6$

So the equation has two real solutions.

A quadratic must contain an $x^2$ term. The equation $2x - 6 = 0$ has none.

$2x - 6 = 0 \;\Rightarrow\; x = 3$

That is linear, not quadratic. The other three all contain an $x^2$ term.

Set each factor equal to $0$.

$x - 7 = 0 \quad \text{or} \quad x + 1 = 0$

So $x = 7$ or $x = -1$.