Introduction to Quadratics: Practice

Set each factor to $0$ and solve the resulting linear equations.

$2x - 6 = 0 \;\Rightarrow\; x = 3, \qquad x + 5 = 0 \;\Rightarrow\; x = -5$

So $x = 3$ or $x = -5$.

Test the candidates by substituting into $2x^2 - 5x - 3$ and looking for $0$.

$2(3)^2 - 5(3) - 3 = 18 - 15 - 3 = 0$

So $x = 3$ is a solution. The other values give nonzero results.

Set each factor equal to $0$.

$3x + 2 = 0 \;\Rightarrow\; x = -\tfrac{2}{3}, \qquad x - 4 = 0 \;\Rightarrow\; x = 4$

So $x = -\tfrac{2}{3}$ or $x = 4$.

Take the square root of both sides. Since $+0$ and $-0$ are the same, the $\pm$ collapses.

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

So there is exactly one solution, $x = 3$.

Isolate the squared term by subtracting $9$ from both sides.

$x^2 = -9$

No real number squares to a negative number, so there is no real solution.

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

$x - 4 = \pm 7$

Then $x = 4 + 7 = 11$ or $x = 4 - 7 = -3$.

A square means the factor multiplied by itself. Multiply it out.

$(x + 4)^2 = (x + 4)(x + 4) = x^2 + 4x + 4x + 16 = x^2 + 8x + 16$

The product $x(x - 7)$ is already factored, with factors $x$ and $x - 7$. Set each to $0$.

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

So $x = 0$ or $x = 7$. It is easy to lose the root $x = 0$ from the first factor.

First isolate the squared term by adding $20$ to both sides.

$x^2 = 25 \;\Rightarrow\; x = \pm 5$

So $x = 5$ or $x = -5$.

Multiply each term of the first factor by each term of the second, then combine like terms.

$(x - 4)(x + 3) = x^2 + 3x - 4x - 12 = x^2 - x - 12$

So the standard form is $x^2 - x - 12 = 0$. The middle term is $-x$ because $3x - 4x = -x$.

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

$x + 1 = \pm 2$

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

If $a = 0$, the squared term $ax^2$ becomes $0$ and vanishes.

$0 \cdot x^2 + bx + c = bx + c = 0$

That is a linear equation, so $a \ne 0$ is exactly what makes an equation genuinely quadratic.