Practice questions

Remove the parentheses. The minus sign in front of ${(x+5)}$ changes both of its signs:

$\begin{align*}8x-12-x-5&=3x+3\\7x-17&=3x+3\end{align*}$

Subtract ${3x}$ from both sides, and add ${17}$:

$\begin{align*}4x&=20\\x&=5\end{align*}$

Check: The left side is ${4(7)-10=18}$. The right side is ${3(5)+3=18}$.

The LCD of ${3}$ and ${2}$ is ${6}$. Multiply every term by ${6}$:

$\begin{align*}2(2x+1)-3(x-2)&=6\\4x+2-3x+6&=6\\x+8&=6\\x&=-2\end{align*}$

Check: ${\dfrac{-3}{3}-\dfrac{-4}{2}=-1+2=1}$. It checks.

Multiply both sides by ${3}$ to clear the fraction. Then divide both sides by ${\pi r^2}$:

$\begin{align*}3V&=\pi r^2h\\h&=\dfrac{3V}{\pi r^2}\end{align*}$

The cars move apart in opposite directions. So the distance between them is the sum of the distances they have each driven.

Let ${t}$ be the time, in hours. Using ${d=rt}$, the cars have driven ${55t}$ and ${65t}$ miles. Their total is ${300}$:

$\begin{align*}55t+65t&=300\\120t&=300\\t&=2.5\end{align*}$

They are ${300}$ miles apart after ${2.5}$ hours.

Check: ${55(2.5)+65(2.5)=137.5+162.5=300}$.

Get ${0}$ on the right side, and factor (by the ${ac}$ method):

$\begin{align*}3x^2-5x-2&=0\\(3x+1)(x-2)&=0\end{align*}$

So ${x=-\dfrac{1}{3}}$ or ${x=2}$.

Take square roots of both sides. Keep both signs, and simplify ${\sqrt{18}=\sqrt{9}\,\sqrt{2}=3\sqrt{2}}$:

${2x+1=\pm 3\sqrt{2}}$

Subtract ${1}$, and divide by ${2}$:

${x=\dfrac{-1\pm 3\sqrt{2}}{2}}$

Move the constant to the right side:

${x^2-8x=-5}$

Half of ${-8}$ is ${-4}$, and ${(-4)^2=16}$. Add ${16}$ to both sides:

$\begin{align*}x^2-8x+16&=-5+16\\(x-4)^2&=11\end{align*}$

Take square roots, then add ${4}$:

$\begin{align*}x-4&=\pm\sqrt{11}\\x&=4\pm\sqrt{11}\end{align*}$

Here

${a=2,\quad b=3,\quad c=-4}$.

Put the values into the formula:

$\begin{align*}&x\\&=\dfrac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)}\\&=\dfrac{-3\pm\sqrt{9+32}}{4}\\&=\dfrac{-3\pm\sqrt{41}}{4}\end{align*}$

As decimals, the solutions are about ${0.85}$ and ${-2.35}$.

Find the discriminant, with ${a=9}$, ${b=-6}$, ${c=1}$:

$\begin{align*}&b^2-4ac\\&=(-6)^2-4(9)(1)\\&=36-36=0\end{align*}$

The discriminant is ${0}$, so there is exactly one real solution. (In fact, ${9x^2-6x+1=(3x-1)^2}$, so ${x=\dfrac{1}{3}}$.)

Use FOIL. Then use ${i^2=-1}$, so ${-15i^2=+15}$:

$\begin{align*}&(4-3i)(2+5i)\\&=8+20i-6i-15i^2\\&=8+14i+15\\&=23+14i\end{align*}$

Multiply the numerator and the denominator by the conjugate of the denominator, ${3+4i}$. The denominator becomes ${3^2+4^2=25}$:

$\begin{align*}&\dfrac{2+i}{3-4i}\cdot\dfrac{3+4i}{3+4i}\\&=\dfrac{6+8i+3i+4i^2}{25}\\&=\dfrac{2+11i}{25}\end{align*}$

Split it into real and imaginary parts:

$\dfrac{2+11i}{25}=\dfrac{2}{25}+\dfrac{11}{25}i$

${42=4\cdot 10+2}$, so the remainder is ${2}$. Use ${i^4=1}$:

$i^{42}=\left(i^4\right)^{10}\cdot i^2=1\cdot(-1)=-1$

So ${i^{42}=-1+0i}$.

Use the quadratic formula with ${a=1}$, ${b=6}$, ${c=13}$:

$\begin{align*}&x\\&=\dfrac{-6\pm\sqrt{6^2-4(1)(13)}}{2(1)}\\&=\dfrac{-6\pm\sqrt{-16}}{2}\\&=\dfrac{-6\pm 4i}{2}\\&=-3\pm 2i\end{align*}$

Factor the last denominator: ${x^2-1=(x-1)(x+1)}$. The denominators are ${0}$ when ${x=1}$ or ${x=-1}$, so these cannot be solutions.

The LCD is ${(x-1)(x+1)}$. Multiply every term by it:

$\begin{align*}2(x+1)-(x-1)&=4\\2x+2-x+1&=4\\x+3&=4\\x&=1\end{align*}$

But ${x=1}$ makes the denominators ${0}$. It is extraneous. So the equation has no solution.

The radical is alone. Square both sides. On the right, use ${(a-b)^2=a^2-2ab+b^2}$:

$\begin{align*}x+7&=x^2-10x+25\\0&=x^2-11x+18\\0&=(x-2)(x-9)\end{align*}$

So ${x=2}$ or ${x=9}$. Check each one in the original equation:

• ${x=9}$: ${\sqrt{16}=4}$, and ${9-5=4}$. True.
• ${x=2}$: ${\sqrt{9}=3}$, but ${2-5=-3}$. False.

So ${2}$ is extraneous. The only solution is ${x=9}$.

The expression ${3x-4}$ is ${8}$ or ${-8}$:

${3x-4=8}$   or   ${3x-4=-8}$

${3x=12}$   or   ${3x=-4}$

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

Let ${u=x^2}$. The equation becomes

$\begin{align*}u^2-13u+36&=0\\(u-4)(u-9)&=0\end{align*}$

So ${x^2=4}$ or ${x^2=9}$. There are four solutions: ${x=\pm 2}$ and ${x=\pm 3}$.

Factor by grouping:

$\begin{align*}x^2(2x-1)-9(2x-1)&=0\\(2x-1)(x^2-9)&=0\\(2x-1)(x+3)(x-3)&=0\end{align*}$

So ${x=\dfrac{1}{2}}$, ${x=-3}$, or ${x=3}$.

Subtract ${1}$ from all three parts:

${-4\le -2x<4}$

Divide all three parts by ${-2}$. The divisor is negative, so both signs reverse:

${2\ge x>-2}$

That is, ${-2<x\le 2}$. In interval notation, the answer is ${(-2,2]}$.

${x+2}$ is more than ${3}$ units from ${0}$:

${x+2<-3}$   or   ${x+2>3}$

${x<-5}$   or   ${x>1}$

The answer is ${(-\infty,-5)\cup(1,\infty)}$.

Factor: ${(x+4)(x-2)\le 0}$. The critical numbers are ${-4}$ and ${2}$.

Interval	Sign of ${(x+4)(x-2)}$
${(-\infty,-4)}$	${(-)(-)=+}$
${(-4,2)}$	${(+)(-)=-}$
${(2,\infty)}$	${(+)(+)=+}$

We need ${\le 0}$. Take the negative interval, and include the critical numbers:

${[-4,2]}$

The numerator is ${0}$ at ${x=3}$, and the denominator is ${0}$ at ${x=-1}$.

Interval	Sign of ${\dfrac{x-3}{x+1}}$
${(-\infty,-1)}$	${\dfrac{-}{-}=+}$
${(-1,3)}$	${\dfrac{-}{+}=-}$
${(3,\infty)}$	${\dfrac{+}{+}=+}$

We need ${>0}$. Take the positive intervals. Neither critical number is included: at ${3}$ the fraction is ${0}$, and at ${-1}$ it is undefined.

${(-\infty,-1)\cup(3,\infty)}$