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Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 2x^2-8x+1=0$

Possible Answers:

$x=8.57\text{ and }x=6.43$ $\displaystyle x=8.57\text{ and }x=6.43$

$x=2.91\text{ and }x=0.09$ $\displaystyle x=2.91\text{ and }x=0.09$

$x=3.87\text{ and }x=0.13$ $\displaystyle x=3.87\text{ and }x=0.13$

$x=1.37\text{ and }x=5.63$ $\displaystyle x=1.37\text{ and }x=5.63$

Correct answer:

$x=3.87\text{ and }x=0.13$ $\displaystyle x=3.87\text{ and }x=0.13$

Explanation:

$\displaystyle 2x^2-8x+1=0$

Start by subtracting $\displaystyle 1$ from both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 2x^2-8x=-1$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

$x^2-4x=-\frac{1}{2}$ $\displaystyle x^2-4x=-\frac{1}{2}$

Now, look at the coefficient of the $\displaystyle x$-term. To complete the square, divide this coefficient by $\displaystyle 2$, then square the result. Add this term to both sides of the equation.

$x^2-4x+4=-\frac{1}{2}+4$ $\displaystyle x^2-4x+4=-\frac{1}{2}+4$

Rewrite the left side of the equation in the squared form.

$(x-2)^2=\frac{7}{2}$ $\displaystyle (x-2)^2=\frac{7}{2}$

Take the square root of both sides.

$x-2=\pm \sqrt{\frac{7}{2}}$ $\displaystyle x-2=\pm \sqrt{\frac{7}{2}}$

Now solve for $\displaystyle x$.

$x=2+\sqrt{\frac{7}{2}}=3.87$ $\displaystyle x=2+\sqrt{\frac{7}{2}}=3.87$

$x=5-\sqrt{\frac{7}{2}}=0.13$ $\displaystyle x=5-\sqrt{\frac{7}{2}}=0.13$

Round to two places after the decimal.

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Example Question #62 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 3x^2-5x-10=0$

Possible Answers:

$x=2.84\text{ and }x=-1.17$ $\displaystyle x=2.84\text{ and }x=-1.17$

$x=3.69\text{ and }x=-2.31$ $\displaystyle x=3.69\text{ and }x=-2.31$

$x=7.08\text{ and }x=-3.92$ $\displaystyle x=7.08\text{ and }x=-3.92$

$x=4.09\text{ and }x=1.91$ $\displaystyle x=4.09\text{ and }x=1.91$

Correct answer:

$x=2.84\text{ and }x=-1.17$ $\displaystyle x=2.84\text{ and }x=-1.17$

Explanation:

$\displaystyle 3x^2-5x-10=0$

Start by adding $\displaystyle 10$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 3x^2-5x=10$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

$x^2-\frac{5}{3}x=\frac{10}{3}$ $\displaystyle x^2-\frac{5}{3}x=\frac{10}{3}$

$x^2-\frac{5}{3}x+\frac{25}{36}=\frac{10}{3}+\frac{25}{36}$ $\displaystyle x^2-\frac{5}{3}x+\frac{25}{36}=\frac{10}{3}+\frac{25}{36}$

Rewrite the left side of the equation in the squared form.

$(x-\frac{5}{6})^2=\frac{145}{36}$ $\displaystyle (x-\frac{5}{6})^2=\frac{145}{36}$

Take the square root of both sides.

$x-\frac{5}{6}=\pm \sqrt{\frac{145}{36}}$ $\displaystyle x-\frac{5}{6}=\pm \sqrt{\frac{145}{36}}$

Now solve for $\displaystyle x$.

$x=\frac{5}{6}+\sqrt{\frac{145}{36}}=2.84$ $\displaystyle x=\frac{5}{6}+\sqrt{\frac{145}{36}}=2.84$

$x=\frac{5}{6}-\sqrt{\frac{145}{36}}=-1.17$ $\displaystyle x=\frac{5}{6}-\sqrt{\frac{145}{36}}=-1.17$

Round to two places after the decimal.

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Example Question #63 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2-10x-12=0$

Possible Answers:

$x=12.36\text{ and }x=2.36$ $\displaystyle x=12.36\text{ and }x=2.36$

$x=-3.46\text{ and }x=-16.46$ $\displaystyle x=-3.46\text{ and }x=-16.46$

$x=20.75\text{ and }x=8.75$ $\displaystyle x=20.75\text{ and }x=8.75$

$x=11.08\text{ and }x=-1.08$ $\displaystyle x=11.08\text{ and }x=-1.08$

Correct answer:

$x=11.08\text{ and }x=-1.08$ $\displaystyle x=11.08\text{ and }x=-1.08$

Explanation:

$\displaystyle x^2-10x-12=0$

Start by adding $\displaystyle 12$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2-10x=12$

$\displaystyle x^2-10x+25=12+25$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x-5)^2=37$

Take the square root of both sides.

$x-5=\pm \sqrt{37}$ $\displaystyle x-5=\pm \sqrt{37}$

Now solve for $\displaystyle x$.

$x=5+\sqrt{37}=11.08$ $\displaystyle x=5+\sqrt{37}=11.08$

$x=5-\sqrt{37}=-1.08$ $\displaystyle x=5-\sqrt{37}=-1.08$

Round to two places after the decimal.

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Example Question #64 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2+18x-20=0$

Possible Answers:

$x=3.39\text{ and }x=-18.39$ $\displaystyle x=3.39\text{ and }x=-18.39$

$x=1.05\text{ and }x=-19.05$ $\displaystyle x=1.05\text{ and }x=-19.05$

$x=9.21\text{ and }x=1.21$ $\displaystyle x=9.21\text{ and }x=1.21$

$x=6.12\text{ and }x=-0.12$ $\displaystyle x=6.12\text{ and }x=-0.12$

Correct answer:

$x=1.05\text{ and }x=-19.05$ $\displaystyle x=1.05\text{ and }x=-19.05$

Explanation:

$\displaystyle x^2+18x-20=0$

Start by adding $\displaystyle 20$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2+18x=20$

$\displaystyle x^2+18x+81=20+81$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x+9)^2=101$

Take the square root of both sides.

$x+9=\pm \sqrt{101}$ $\displaystyle x+9=\pm \sqrt{101}$

Now solve for $\displaystyle x$.

$x=-9+\sqrt{101}=1.05$ $\displaystyle x=-9+\sqrt{101}=1.05$

$x=-9-\sqrt{101}=-19.05$ $\displaystyle x=-9-\sqrt{101}=-19.05$

Round to two places after the decimal.

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Example Question #65 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 8x^2-2x-1=0$

Possible Answers:

$x=\frac{1}{8}\text{ and }x=-\frac{1}{2}$ $\displaystyle x=\frac{1}{8}\text{ and }x=-\frac{1}{2}$

$x=\frac{5}{8}\text{ and }x=-\frac{7}{8}$ $\displaystyle x=\frac{5}{8}\text{ and }x=-\frac{7}{8}$

$x=\frac{1}{4}\text{ and }x=-\frac{3}{8}$ $\displaystyle x=\frac{1}{4}\text{ and }x=-\frac{3}{8}$

$x=\frac{1}{2}\text{ and }x=-\frac{1}{4}$ $\displaystyle x=\frac{1}{2}\text{ and }x=-\frac{1}{4}$

Correct answer:

$x=\frac{1}{2}\text{ and }x=-\frac{1}{4}$ $\displaystyle x=\frac{1}{2}\text{ and }x=-\frac{1}{4}$

Explanation:

$\displaystyle 8x^2-2x-1=0$

Start by adding $\displaystyle 1$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 8x^2-2x=1$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

$x^2-\frac{1}{4}x=\frac{1}{8}$ $\displaystyle x^2-\frac{1}{4}x=\frac{1}{8}$

$x^2-\frac{1}{4}x+\frac{1}{64}=\frac{1}{8}+\frac{1}{64}$ $\displaystyle x^2-\frac{1}{4}x+\frac{1}{64}=\frac{1}{8}+\frac{1}{64}$

Rewrite the left side of the equation in the squared form.

$(x-\frac{1}{8})^2=\frac{9}{64}$ $\displaystyle (x-\frac{1}{8})^2=\frac{9}{64}$

Take the square root of both sides.

$x-\frac{1}{8}=\pm \sqrt{\frac{9}{64}}$ $\displaystyle x-\frac{1}{8}=\pm \sqrt{\frac{9}{64}}$

Now solve for $\displaystyle x$.

$x=\frac{1}{8}+\frac{3}{8}=\frac{1}{2}$ $\displaystyle x=\frac{1}{8}+\frac{3}{8}=\frac{1}{2}$

$x=\frac{1}{8}-\frac{3}{8}=-\frac{1}{4}$ $\displaystyle x=\frac{1}{8}-\frac{3}{8}=-\frac{1}{4}$

Round to two places after the decimal.

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Example Question #66 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2+16x+50=0$

Possible Answers:

$x=-1.51\text{ and }x=-10.49$ $\displaystyle x=-1.51\text{ and }x=-10.49$

$x=-3.18\text{ and }x=-8.82$ $\displaystyle x=-3.18\text{ and }x=-8.82$

$x=-6.68\text{ and }x=-10.32$ $\displaystyle x=-6.68\text{ and }x=-10.32$

$x=-4.26\text{ and }x=-11.74$ $\displaystyle x=-4.26\text{ and }x=-11.74$

Correct answer:

$x=-4.26\text{ and }x=-11.74$ $\displaystyle x=-4.26\text{ and }x=-11.74$

Explanation:

$\displaystyle x^2+16x+50=0$

Start by subtracting $\displaystyle 50$ from both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2+16x=-50$

$\displaystyle x^2+16x+64=-50+64$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x+8)^2=14$

Take the square root of both sides.

$x+8=\pm \sqrt{14}$ $\displaystyle x+8=\pm \sqrt{14}$

Now solve for $\displaystyle x$.

$x=-8+\sqrt{14}=-4.26$ $\displaystyle x=-8+\sqrt{14}=-4.26$

$x=-8+\sqrt{14}=-11.74$ $\displaystyle x=-8+\sqrt{14}=-11.74$

Round to two places after the decimal.

Report an Error

Example Question #67 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2+20x+50=0$

Possible Answers:

$x=-3.07\text{ and }x=-18.93$ $\displaystyle x=-3.07\text{ and }x=-18.93$

$x=-2.93\text{ and }x=-17.07$ $\displaystyle x=-2.93\text{ and }x=-17.07$

$x=-6.27\text{ and }x=-18.73$ $\displaystyle x=-6.27\text{ and }x=-18.73$

$x=-4.73\text{ and }x=-20.27$ $\displaystyle x=-4.73\text{ and }x=-20.27$

Correct answer:

$x=-2.93\text{ and }x=-17.07$ $\displaystyle x=-2.93\text{ and }x=-17.07$

Explanation:

$\displaystyle x^2+20x+50=0$

Start by subtracting $\displaystyle 50$ from both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2+20x=-50$

$\displaystyle x^2+20x+100=-50+100$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x+10)^2=50$

Take the square root of both sides.

$x+10=\pm \sqrt{50}$ $\displaystyle x+10=\pm \sqrt{50}$

Now solve for $\displaystyle x$.

$x=-10+\sqrt{50}=-2.93$ $\displaystyle x=-10+\sqrt{50}=-2.93$

$x=-10+\sqrt{50}=-17.07$ $\displaystyle x=-10+\sqrt{50}=-17.07$

Round to two places after the decimal.

Report an Error

Example Question #68 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 3x^2+4x-6=0$

Possible Answers:

$x=6.31\text{ and }x=10.81$ $\displaystyle x=6.31\text{ and }x=10.81$

$x=-9.37\text{ and }x=5.08$ $\displaystyle x=-9.37\text{ and }x=5.08$

$x=2.22\text{ and }x=-8.09$ $\displaystyle x=2.22\text{ and }x=-8.09$

$x=0.90\text{ and }x=-2.23$ $\displaystyle x=0.90\text{ and }x=-2.23$

Correct answer:

$x=0.90\text{ and }x=-2.23$ $\displaystyle x=0.90\text{ and }x=-2.23$

Explanation:

$\displaystyle 3x^2+4x-6=0$

Start by adding $\displaystyle 6$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 3x^2+4x=6$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

$x^2+\frac{4}{3}x=2$ $\displaystyle x^2+\frac{4}{3}x=2$

$x^2+\frac{4}{3}x+\frac{4}{9}=2+\frac{4}{9}$ $\displaystyle x^2+\frac{4}{3}x+\frac{4}{9}=2+\frac{4}{9}$

Rewrite the left side of the equation in the squared form.

$(x+\frac{2}{3})^2=\frac{22}{9}$ $\displaystyle (x+\frac{2}{3})^2=\frac{22}{9}$

Take the square root of both sides.

$x+\frac{2}{3}=\pm \sqrt{\frac{22}{9}}$ $\displaystyle x+\frac{2}{3}=\pm \sqrt{\frac{22}{9}}$

Now solve for $\displaystyle x$.

$x=-\frac{2}{3}+\sqrt{\frac{22}{9}}=0.90$ $\displaystyle x=-\frac{2}{3}+\sqrt{\frac{22}{9}}=0.90$

$x=-\frac{2}{3}-\sqrt{\frac{22}{9}}=-2.23$ $\displaystyle x=-\frac{2}{3}-\sqrt{\frac{22}{9}}=-2.23$

Round to two places after the decimal.

Report an Error

Example Question #69 : Completing The Square

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 10x^2-5x-6=0$

Possible Answers:

$x=3.51\text{ and }x=0.51$ $\displaystyle x=3.51\text{ and }x=0.51$

$x=27.66\text{ and }x=-2.66$ $\displaystyle x=27.66\text{ and }x=-2.66$

$x=1.06\text{ and }x=-0.56$ $\displaystyle x=1.06\text{ and }x=-0.56$

$x=2.95\text{ and }x=1.05$ $\displaystyle x=2.95\text{ and }x=1.05$

Correct answer:

$x=1.06\text{ and }x=-0.56$ $\displaystyle x=1.06\text{ and }x=-0.56$

Explanation:

$\displaystyle 10x^2-5x-6=0$

Start by adding $\displaystyle 6$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 10x^2-5x=6$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

$x^2-\frac{1}{2}x=\frac{3}{5}$ $\displaystyle x^2-\frac{1}{2}x=\frac{3}{5}$

$x^2-\frac{1}{2}x+\frac{1}{16}=\frac{3}{5}+\frac{1}{16}$ $\displaystyle x^2-\frac{1}{2}x+\frac{1}{16}=\frac{3}{5}+\frac{1}{16}$

Rewrite the left side of the equation in the squared form.

$(x-\frac{1}{4})^2=\frac{53}{80}$ $\displaystyle (x-\frac{1}{4})^2=\frac{53}{80}$

Take the square root of both sides.

$x-\frac{1}{4}=\pm \sqrt{\frac{53}{80}}$ $\displaystyle x-\frac{1}{4}=\pm \sqrt{\frac{53}{80}}$

Now solve for $\displaystyle x$.

$x=\frac{1}{4}+\sqrt{\frac{53}{80}}=1.06$ $\displaystyle x=\frac{1}{4}+\sqrt{\frac{53}{80}}=1.06$

$x=\frac{1}{4}-\sqrt{\frac{53}{80}}=-0.56$ $\displaystyle x=\frac{1}{4}-\sqrt{\frac{53}{80}}=-0.56$

Round to two places after the decimal.

Report an Error

Example Question #70 : Completing The Square

Which number completes the following square equation:

$x^{2}+6x+$ $\displaystyle x^{2}+6x+$___

Possible Answers:

$\displaystyle 36$

$\displaystyle 9$

$\displaystyle 3$

$\displaystyle 6$

$\displaystyle 12$

Correct answer:

$\displaystyle 9$

Explanation:

In order to complete the square, you half the middle number and square it.

$x^{2}+6x+$ $\displaystyle x^{2}+6x+$__________

$\downarrow$ $\displaystyle \downarrow$ $\uparrow$ $\displaystyle \uparrow$

$\left(\frac{6}{2}\right)^2=(3)^2=9$ $\displaystyle \left(\frac{6}{2}\right)^2=(3)^2=9$

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Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #61 : Completing The Square

Example Question #62 : Completing The Square

Example Question #63 : Completing The Square

Example Question #64 : Completing The Square

Example Question #65 : Completing The Square

Example Question #66 : Completing The Square

Example Question #67 : Completing The Square

Example Question #68 : Completing The Square

Example Question #69 : Completing The Square

Example Question #70 : Completing The Square

All Algebra II Resources

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