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Algebra II : Algebra II

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 #3 : Completing The Square

Solve the following equation by completing the square. Use a calculator to determine the answer to the closest hundredth.

$\displaystyle x^2 + 8x - 12 = 0$

Possible Answers:

1.29 $\displaystyle 1.29$ and -9.29 $\displaystyle -9.29$

12.45 $\displaystyle 12.45$ and -10.45 $\displaystyle -10.45$

$\displaystyle 2$ and $\displaystyle 6$

2.38 $\displaystyle 2.38$ and -3.38 $\displaystyle -3.38$

$\displaystyle -2$ and $\displaystyle 6$

Correct answer:

1.29 $\displaystyle 1.29$ and -9.29 $\displaystyle -9.29$

Explanation:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 8x - 12 = 0$

$\displaystyle x^2 + 8x = 12$

Then, divide the middle coefficient by 2:

$\frac{8}{2} = 4$ $\displaystyle \frac{8}{2} = 4$

Square that and add it to both sides:

$\displaystyle 4^2 = 16$

$\displaystyle x^2 + 8x +16 = 12 +16$

$\displaystyle x^2 + 8x + 16 = 28$

Now, you can easily factor the quadratic:

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

$\displaystyle (x+4)^2=28$

Take the square root of both sides:

$x+ 4 = \pm\sqrt{28}$ $\displaystyle x+ 4 = \pm\sqrt{28}$

Finish out the solution:

$x=\pm\sqrt{28}-4$ $\displaystyle x=\pm\sqrt{28}-4$

$\sqrt{28} - 4 = 1.29150262212918$ $\displaystyle \sqrt{28} - 4 = 1.29150262212918$

$-\sqrt{28} - 4 = -9.29150262212918$ $\displaystyle -\sqrt{28} - 4 = -9.29150262212918$

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Example Question #4 : Completing The Square

Solve the following equation by completing the square. Use a calculator to determine the answer to the closest hundredth.

$\displaystyle x^2 - 12x - 16 = 0$

Possible Answers:

13.21 $\displaystyle 13.21$ and -1.21 $\displaystyle -1.21$

16.23 $\displaystyle 16.23$ and 2.23 $\displaystyle 2.23$

23.73 $\displaystyle 23.73$ and -12.73 $\displaystyle -12.73$

No solution

15.72 $\displaystyle 15.72$ and -2.72 $\displaystyle -2.72$

Correct answer:

13.21 $\displaystyle 13.21$ and -1.21 $\displaystyle -1.21$

Explanation:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 - 12x - 16 = 0$

$\displaystyle x^2 - 12x = 16$

Then, divide the middle coefficient by 2:

$\frac{-12}{2}=-6$ $\displaystyle \frac{-12}{2}=-6$

Square that and add it to both sides:

$\displaystyle (-6)^2 = 36$

$\displaystyle x^2 - 12x +36 = 16 + 36$

$\displaystyle x^2 - 12x + 36 = 52$

Now, you can easily factor the quadratic:

$\displaystyle x^2 - 12x + 36 = (x-6)^2$

$\displaystyle (x-6)^2=52$

Take the square root of both sides:

$x- 6 = \pm\sqrt{52}$ $\displaystyle x- 6 = \pm\sqrt{52}$

Finish out the solution:

$x=\pm\sqrt{52}+6$ $\displaystyle x=\pm\sqrt{52}+6$

$\sqrt{52} + 6 = 13.21110255092798$ $\displaystyle \sqrt{52} + 6 = 13.21110255092798$

$-\sqrt{52} + 6 = -1.21110255092798$ $\displaystyle -\sqrt{52} + 6 = -1.21110255092798$

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Example Question #1 : Completing The Square

Solve the following equation by completing the square. Use a calculator to determine the answer to the closest hundredth.

$\displaystyle x^2 - 20x + 31 = 0$

Possible Answers:

16.73 $\displaystyle 16.73$ and 1.38 $\displaystyle 1.38$

12.38 $\displaystyle 12.38$ and -4.38 $\displaystyle -4.38$

18.31 $\displaystyle 18.31$ and 1.69 $\displaystyle 1.69$

No solution

-18.83 $\displaystyle -18.83$ and 7.23 $\displaystyle 7.23$

Correct answer:

18.31 $\displaystyle 18.31$ and 1.69 $\displaystyle 1.69$

Explanation:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 - 20x + 31 = 0$

$\displaystyle x^2 - 20x =- 31$

Then, divide the middle coefficient by 2:

$\frac{-20}{2}=-10$ $\displaystyle \frac{-20}{2}=-10$

Square that and add it to both sides:

$\displaystyle (-10)^2 = 100$

$\displaystyle x^2 - 20x +100 =- 31 + 100$

$\displaystyle x^2 - 20x +100 = 69$

Now, you can easily factor the quadratic:

$\displaystyle x^2 - 20x +100 = (x - 10)^2$

$\displaystyle (x-10)^2=69$

Take the square root of both sides:

$x - 10 = \pm\sqrt{69}$ $\displaystyle x - 10 = \pm\sqrt{69}$

Finish out the solution:

$x=\pm\sqrt{69}+10$ $\displaystyle x=\pm\sqrt{69}+10$

$\sqrt{69} +10 = 18.30662386291807$ $\displaystyle \sqrt{69} +10 = 18.30662386291807$

$-\sqrt{69} +10 = 1.69337613708192$ $\displaystyle -\sqrt{69} +10 = 1.69337613708192$

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Example Question #6 : Completing The Square

Solve the following equation by completing the square. Use a calculator to determine the answer to the closest hundredth.

$\displaystyle x^2 + 5x - 12 = 0$

Possible Answers:

-6.77 $\displaystyle -6.77$ and 1.77 $\displaystyle 1.77$

No solution

18.38 $\displaystyle 18.38$ and 21.33 $\displaystyle 21.33$

1.38 $\displaystyle 1.38$ and 7.38 $\displaystyle 7.38$

-4.52 $\displaystyle -4.52$ and 3.73 $\displaystyle 3.73$

Correct answer:

-6.77 $\displaystyle -6.77$ and 1.77 $\displaystyle 1.77$

Explanation:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 5x - 12 = 0$

$\displaystyle x^2 + 5x = 12$

Then, divide the middle coefficient by 2:

$\frac{5}{2}= 2.5$ $\displaystyle \frac{5}{2}= 2.5$

Square that and add it to both sides:

$\displaystyle (2.5)^2 = 6.25$

$\displaystyle x^2 + 5x + 6.25 = 12 + 6.25$

$\displaystyle x^2 + 5x + 6.25 = 18.25$

Now, you can easily factor the quadratic:

$\displaystyle x^2 + 5x + 6.25 = (x+2.5)^2$

$\displaystyle (x+2.5)^2=18.25$

Take the square root of both sides:

$x + 2.5 = \pm \sqrt{18.25}$ $\displaystyle x + 2.5 = \pm \sqrt{18.25}$

Finish out the solution:

$x=\pm\sqrt{18.25}-2.5$ $\displaystyle x=\pm\sqrt{18.25}-2.5$

$\sqrt{18.25} - 2.5=1.77200187265877$ $\displaystyle \sqrt{18.25} - 2.5=1.77200187265877$

$-\sqrt{18.25} - 2.5 = -6.77200187265876$ $\displaystyle -\sqrt{18.25} - 2.5 = -6.77200187265876$

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Example Question #2 : Completing The Square

Solve the following equation by completing the square. Use a calculator to determine the answer to the closest hundredth.

$\displaystyle x^2 + 7x + 15 = 0$

Possible Answers:

No solution

12.13 $\displaystyle 12.13$ and -2.13 $\displaystyle -2.13$

-5.48 $\displaystyle -5.48$ and 7.32 $\displaystyle 7.32$

12.33 $\displaystyle 12.33$ and 15.23 $\displaystyle 15.23$

3.47 $\displaystyle 3.47$ and -3.47 $\displaystyle -3.47$

Correct answer:

No solution

Explanation:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 7x + 15 = 0$

$\displaystyle x^2 + 7x = - 15$

Then, divide the middle coefficient by 2:

$\frac{7}{2} = 3.5$ $\displaystyle \frac{7}{2} = 3.5$

Square that and add it to both sides:

$\displaystyle (3.5)^2 = 12.25$

$\displaystyle x^2 + 7x + 12.25 = - 15 + 12.25$

$\displaystyle x^2 + 7x + 12.25 = -2.75$

Now, you can easily factor the quadratic:

$\displaystyle x^2 + 7x + 12.25 = (x+3.5)^2$

$\displaystyle (x+3.5)^2=-2.75$

Your next step would be to take the square root of both sides. At this point, however, you know that you cannot solve the problem. When you take the square root of both sides, you will be forced to take the square root of -2.75 $\displaystyle -2.75$. This is impossible (at least in terms of real numbers), meaning that this problem must have no real solution.

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Example Question #7 : Completing The Square

Use completing the square to solve the following equation, simplifying radicals completely:
$\displaystyle 4 = 2x^2 + 12x -18$

Possible Answers:

$x = -1 \pm 2\sqrt{5}$ $\displaystyle x = -1 \pm 2\sqrt{5}$

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

$x = -3 \pm3\sqrt{2}$ $\displaystyle x = -3 \pm3\sqrt{2}$

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

$x = -3 \pm \sqrt{11}$ $\displaystyle x = -3 \pm \sqrt{11}$

Correct answer:

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

Explanation:

From the original equation, we add 18 to both sides in order to set up our "completing the square."
$\displaystyle 4 = 2x^2 + 12x -18$

$\displaystyle 4+18 = 2x^2 +12x - 18 + 18$

$\displaystyle 22 = 2x^2 + 12x$

To make completing the square sensible, we divide both sides by 2.

$\frac{22}{2} = \frac{2x^2 + 12x}{2}$ $\displaystyle \frac{22}{2} = \frac{2x^2 + 12x}{2}$

$\displaystyle 11 = x^2 + 6x$

We now divide the x coefficient by 2, square the result, and add that to both sides.

$\frac{6}{2} = 3$ $\displaystyle \frac{6}{2} = 3$

3^2 = 9 $\displaystyle 3^2 = 9$

$\displaystyle 11 + 9 = x^2 + 6x + 9$

$\displaystyle 20 = x^2 + 6x + 9$

Since the right side is now a perfect square, we can rewrite it as a square binomial.

$\displaystyle 20 = (x+3)^2$

Take the square root of both sides, simplify the radical and solve for x.

$\sqrt{20}=\sqrt{(x+3)^2}$ $\displaystyle \sqrt{20}=\sqrt{(x+3)^2}$

$\pm2\sqrt{5} = x+ 3$ $\displaystyle \pm2\sqrt{5} = x+ 3$

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

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Example Question #8 : Completing The Square

Solve the following equation by completing the square:
$\displaystyle 2x^2 -4x + 1 = 3$

Possible Answers:

x =1 $\displaystyle x =1$

$x = 1 \pm \sqrt{2}}$ $\displaystyle x = 1 \pm \sqrt{2}}$

$x = -1 \pm 1$ $\displaystyle x = -1 \pm 1$

$x = 1 \pm \sqrt{3}$ $\displaystyle x = 1 \pm \sqrt{3}$

$x = -1 \pm \sqrt{2}$ $\displaystyle x = -1 \pm \sqrt{2}$

Correct answer:

$x = 1 \pm \sqrt{2}}$ $\displaystyle x = 1 \pm \sqrt{2}}$

Explanation:

We start by moving the constant term of the quadratic to the other side of the equation, to set up the "completing the square" format.

$\displaystyle 2x^2 -4x + 1 = 3$

$\displaystyle 2x^2 -4x + 1 - 1 = 3 - 1$

$\displaystyle 2x^2 -4x = 2$

Now to make completing the square sensible, we divide boths sides by 2 so that x^2 will not have a coefficient.

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

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

Now we can complete the square by dividing the x coefficient by 2 and squaring the result, then adding that result to both sides.

$\frac{-2}{2} = -1$ $\displaystyle \frac{-2}{2} = -1$

$\displaystyle (-1)^2 = 1$

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

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

Because the left side is now a perfect square, we can rewrite it as a squared binomial.

$\displaystyle (x-1)^2 = 2$

Take the square root of both sides, and then solve for x.

$\sqrt{(x-1)^2} = \sqrt{2}$ $\displaystyle \sqrt{(x-1)^2} = \sqrt{2}$

$x-1 = \pm \sqrt{2}$ $\displaystyle x-1 = \pm \sqrt{2}$

$x = 1 \pm \sqrt{2}$ $\displaystyle x = 1 \pm \sqrt{2}$

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Example Question #11 : Completing The Square

Solve the following equation by completing the square.

$\displaystyle x^2 - 4x - 16 = 5$

Possible Answers:

$x = 2 \pm\sqrt{21}$ $\displaystyle x = 2 \pm\sqrt{21}$

$x = \left \{ -3, 7\right \}$ $\displaystyle x = \left \{ -3, 7\right \}$

x = 7 $\displaystyle x = 7$

$x = \left \{ -7, 3\right \}$ $\displaystyle x = \left \{ -7, 3\right \}$

$x = 2 \pm\sqrt{29}$ $\displaystyle x = 2 \pm\sqrt{29}$

Correct answer:

$x = \left \{ -3, 7\right \}$ $\displaystyle x = \left \{ -3, 7\right \}$

Explanation:

Starting with the original equation, we move the constant term of the quadratic over to the other side, so we can set up our "completing the square."

$\displaystyle x^2 - 4x - 16 = 5$

$\displaystyle x^2 - 4x - 16 + 16 = 5 + 16$

$\displaystyle x^2 - 4x = 21$

Since the x^2 coefficient is already 1, we don't have to do any division. We can go straight to completing the square by dividing the x coefficient by 2, squaring the result, and adding that result to both sides.

$\frac{-4}{2} = -2$ $\displaystyle \frac{-4}{2} = -2$

$\displaystyle (-2)^2 = 4$

$\displaystyle x^2 - 4x + 4 = 21 + 4$

$\displaystyle x^2 - 4x + 4 = 25$

Since the left side is now a perfect square, we can rewrite it as a squared binomial.

$\displaystyle (x-2)^2 = 25$

Now take the square root of both sides and solve for x.

$\sqrt{(x-2)^2} = \sqrt{25}$ $\displaystyle \sqrt{(x-2)^2} = \sqrt{25}$

$x-2 = \pm 5$ $\displaystyle x-2 = \pm 5$

$x = 2 \pm5$ $\displaystyle x = 2 \pm5$

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Example Question #12 : Completing The Square

Rewrite the follow parabola equation in vertex form:

$\displaystyle y = 4x^2 + 24x + 16$

Possible Answers:

$\displaystyle y = 4(x+3)^2 - 36$

$\displaystyle y = 4(x+3)^2 - 20$

$\displaystyle y = 4(x+3)^2$

$\displaystyle y = (x+3)^2 - 5$

$\displaystyle y = 4(x+3)^2 - 5$

Correct answer:

$\displaystyle y = 4(x+3)^2 - 20$

Explanation:

We start by moving the constant term of the quadratic over to the other side of the equation, in order to set up our "completing the square" form.

$\displaystyle y = 4x^2 + 24x + 16$

$\displaystyle y - 16 = 4x^2 + 24x + 16 -16$

$\displaystyle y -16 = 4x^2 + 24x$

Next we divide both sides by 4 so that the x^2 coefficient will be 1. That will allow us to complete the square.

$\frac{y-16}{4} = \frac{4x^2 + 24x}{4}$ $\displaystyle \frac{y-16}{4} = \frac{4x^2 + 24x}{4}$

$\frac{y}{4} - 4 = x^2 + 6 x$ $\displaystyle \frac{y}{4} - 4 = x^2 + 6 x$

Now we are ready to complete the square. We divide the x coefficient by 2, square the result, and add that result to both sides.

$\frac{6}{2} = 3$ $\displaystyle \frac{6}{2} = 3$

3^2 = 9 $\displaystyle 3^2 = 9$

$\frac{y}{4} - 4 + 9 = x^2 + 6 x + 9$ $\displaystyle \frac{y}{4} - 4 + 9 = x^2 + 6 x + 9$

$\frac{y}{4} +5 = x^2 + 6 x + 9$ $\displaystyle \frac{y}{4} +5 = x^2 + 6 x + 9$

Because the right side is now a perfect square, we can rewrite it as a squared binomial.

$\frac{y}{4} + 5 = (x+3)^2$ $\displaystyle \frac{y}{4} + 5 = (x+3)^2$

To finish, all we have to do now is solve for y. We'll subtract 5 to both sides, then multiply by 4.

$\frac{y}{4} + 5 -5 = (x+3)^2 - 5$ $\displaystyle \frac{y}{4} + 5 -5 = (x+3)^2 - 5$

$\frac{y}{4} = (x+3)^2 - 5$ $\displaystyle \frac{y}{4} = (x+3)^2 - 5$

$4*\frac{y}{4} = ((x+3)^2 - 5)*4$ $\displaystyle 4*\frac{y}{4} = ((x+3)^2 - 5)*4$

$\displaystyle y = 4(x+3)^2 - 5*4$

$\displaystyle y = 4(x+3)^2 - 20$

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Example Question #81 : Solving Quadratic Equations

Solve by completing the square.

$x^{2}-8x+21=6$ $\displaystyle x^{2}-8x+21=6$

Possible Answers:

x=1,5 $\displaystyle x=1,5$

x=1,3 $\displaystyle x=1,3$

x=-5,5 $\displaystyle x=-5,5$

x=3,5 $\displaystyle x=3,5$

x=5 $\displaystyle x=5$

Correct answer:

x=3,5 $\displaystyle x=3,5$

Explanation:

The first step is to make sure the x² term has a coefficient of 1. Since we have that, we move onto the next step.

Next, move the "loose" number over to the other side.

$x^{2}-8x=-15$ $\displaystyle x^{2}-8x=-15$

Now divide the coefficient of the x term by 2 (don't forget the sign!). Add the square of this number to both sides.

$-8x \rightarrow -\frac{8}{2}=-4 \rightarrow (-4)^{2}=16$ $\displaystyle -8x \rightarrow -\frac{8}{2}=-4 \rightarrow (-4)^{2}=16$

$x^{2}-8x+16=-15+16$ $\displaystyle x^{2}-8x+16=-15+16$

Simplify:

$x^{2}-8x+16=1$ $\displaystyle x^{2}-8x+16=1$

Now the left side of the equation can be simplified to a squared factor. The factor that will be squared is going to be x plus the original x coefficient divided by 2 as we calculated above.

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

Take the square root of both sides:

$x-4=\pm 1$ $\displaystyle x-4=\pm 1$

Solve for x:

$x-4=1 \rightarrow x=5$ $\displaystyle x-4=1 \rightarrow x=5$

$x-4=-1 \rightarrow x=3$ $\displaystyle x-4=-1 \rightarrow x=3$

x=3,5 $\displaystyle x=3,5$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3 : Completing The Square

Example Question #4 : Completing The Square

Example Question #1 : Completing The Square

Example Question #6 : Completing The Square

Example Question #2 : Completing The Square

Example Question #7 : Completing The Square

Example Question #8 : Completing The Square

Example Question #11 : Completing The Square

Example Question #12 : Completing The Square

Example Question #81 : Solving Quadratic Equations

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