Introduction to Quadratics: Practice

Take the square root of both sides, keeping both signs, and simplify $\sqrt{12} = 2\sqrt{3}$.

$x - 5 = \pm 2\sqrt{3}$

Then $x = 5 \pm 2\sqrt{3}$, which is two irrational roots. Both signs are kept, so it is not a single value.

Set each factor equal to $0$ and solve.

$2x + 3 = 0 \;\Rightarrow\; x = -\tfrac{3}{2}, \qquad 3x - 6 = 0 \;\Rightarrow\; x = 2$

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

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

$(2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15$

Isolate the square: $2x^2 = -8$, so $x^2 = -4$.

$x^2 = -4 < 0$

No real number squares to a negative, so there are no real solutions.

A single solution happens when a squared factor equals $0$, giving a repeated root.

$(x - 3)^2 = 0 \;\Rightarrow\; x = 3$

Here $x^2 = 9$ has two roots, $x^2 = -9$ has none, and $(x - 3)(x + 3) = 0$ has two.

First isolate the squared group by subtracting $5$ from both sides.

$(x - 1)^2 = 9 \;\Rightarrow\; x - 1 = \pm 3$

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

Set each factor equal to $0$ and solve.

$5x - 10 = 0 \;\Rightarrow\; x = 2, \qquad 2x + 7 = 0 \;\Rightarrow\; x = -\tfrac{7}{2}$

So $x = 2$ or $x = -\tfrac{7}{2}$.

Rearrange each to standard form and check whether an $x^2$ term survives.

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

In the first and third the $x^2$ terms cancel, leaving no squared term, and the second is already linear. Only $x^2 - x = 6$ is quadratic.

Each root $r$ comes from a factor $x - r$. Roots $-1$ and $5$ give factors $x + 1$ and $x - 5$.

$(x + 1)(x - 5) = 0$

Check: $x = -1$ makes the first factor $0$, and $x = 5$ makes the second $0$.

The left side is a square, which is never negative for a real $x$.

$(x + 2)^2 \ge 0 > -1$

So $(x + 2)^2 = -1$ has no real solution.

The expression is already a product of $x$ and $2x - 9$. Set each factor to $0$.

$x = 0 \quad \text{or} \quad 2x - 9 = 0 \;\Rightarrow\; x = \tfrac{9}{2}$

So $x = 0$ or $x = \tfrac{9}{2}$; do not lose the root $x = 0$.

Set each factor to $0$: $3x - 2 = 0$ gives $x = \tfrac{2}{3}$, and $x + 7 = 0$ gives $x = -7$.

$x = \tfrac{2}{3} \quad \text{or} \quad x = -7$

Of the listed choices, only $x = -7$ is a root.