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Use Square Roots to Solve Quadratic Equations

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Solving Quadratics Using Square Roots

Mrs. Garber draws a square on the board and writes the equation \frac{4s^2}{5} - 3 = 13 on the board. "This equation represents the area," she says. "What is the length of each side ( s )?"

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Khan Academy: Solving Quadratics by Square Roots

Guidance

Now that you are familiar with square roots, we will use them to solve quadratic equations. Keep in mind, that square roots cannot be used to solve every type of quadratic. In order to solve a quadratic equation by using square roots, an x- term cannot be present. Solving a quadratic equation by using square roots is very similar to solving a linear equation. In the end, you must isolate the x^2 or whatever is being squared.

Example A

Solve 2x^2-3=15 .

Solution: Start by isolating the x^2 .

2x^2-3 &= 15\\2x^2 &= 18\\x^2 &= 9

At this point, you can take the square root of both sides.

\sqrt{x^2} &= \pm \sqrt{9}\\x &= \pm 3

Notice that x has two solutions; 3 or -3. When taking the square root, always put the \pm (plus or minus sign) in front of the square root. This indicates that the positive or negative answer will be the solution.

Check :

& 2(3)^2-3 = 15 \qquad 2(-3)^2-3 = 15 \\& \ 2 \cdot 9 -3 = 15 \quad or \quad \ 2 \cdot 9 -3 = 15 \\& \quad 18-3 = 15 \qquad \qquad 18-3 = 15

Example B

Solve \frac{x^2}{16}+3=27 .

Solution: Isolate x^2 and then take the square root.

\frac{x^2}{16}+3 &= 27\\\frac{x^2}{16} &= 24\\x^2 &= 384\\x &= \pm \sqrt{384}= \pm 8\sqrt{6}

Example C

Solve 3(x-5)^2+7=43 .

Solution: In this example, x is not the only thing that is squared. Isolate the (x-5)^2 and then take the square root.

3(x-5)^2+7 &= 43\\3(x-5)^2 &= 36\\(x-5)^2 &= 12\\x-5 &= \pm \sqrt{12} \ or \ \pm 2\sqrt{3}

Now that the square root is gone, add 5 to both sides.

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

x=5+2\sqrt{3} or 5-2\sqrt{3} . We can estimate these solutions as decimals; 8.46 or 1.54. Remember, that the most accurate answer includes the radical numbers.

Intro Problem Revisit To find s , isolate s^2 and then take the square root.

\frac{4s^2}{5} - 3 &= 13\\\frac{4s^2}{5} &= 16\\s^2 &= 20\\s &= \pm \sqrt{20}= \pm 2\sqrt{5}

Therefore the length of the square's side is 2\sqrt{5} .

Guided Practice

Solve the following quadratic equations.

1. \frac{2}{3}x^2-14=38

2. 11+x^2=4x^2+5

3. (2x+1)^2-6=19

Answers

1. Isolate x^2 and take the square root.

\frac{2}{3}x^2-14 &= 38\\\frac{2}{3}x^2 &= 52\\x^2 &= 78\\x &= \pm \sqrt{78}

2. Combine all like terms, then isolate x^2 .

11+x^2 &= 4x^2+5\\-3x^2 &= -6\\x^2 &= 2\\x &= \pm \sqrt{2}

3. Isolate what is being squared, take the square root, and then isolate x .

(2x+1)^2-6 &= 19\\(2x+1)^2 &= 25\\2x+1 &= \pm 5\\2x &= -1 \pm 5\\x &= \frac{-1 \pm 5}{2} \rightarrow x = \frac{-1+5}{2}=2 \ or \ x = \frac{-1-5}{2}=-3

Practice

Solve the following quadratic equations. Reduce answers as much as possible. No decimals.

  1. x^2=144
  2. 5x^2-4=16
  3. 8-10x^2=-22
  4. (x+2)^2=49
  5. 6(x-5)^2+1=19
  6. \frac{3}{4}x^2-19=26
  7. x^2-12=36-2x^2
  8. 9-\frac{x^2}{3}=-33
  9. -4(x+7)^2=-52
  10. 2(3x+4)^2-5=45
  11. \frac{1}{3}(x-10)^2-8=16
  12. \frac{(x-1)^2}{6}-\frac{8}{3}=\frac{7}{2}

Use either factoring or solving by square roots to solve the following quadratic equations.

  1. x^2-16x+55=0
  2. 2x^2-9=27
  3. 6x^2+23x=-20
  4. Writing Write a set of hints that will help you remember when you should solve an equation by factoring and by square roots. Are there any quadratics that can be solved using either method?
  5. Solve x^2-9=0 by factoring and by using square roots. Which do you think is easier? Why?
  6. Solve (3x-2)^2+1=17 by using square roots. Then, solve 3x^2-4x-4=0 by factoring. What do you notice? What can you conclude?
  7. Real Life Application The aspect ratio of a TV screen is the ratio of the screen’s width to its height. For HDTVs, the aspect ratio is 16:9. What is the width and height of a 42 inch screen TV? (42 inches refers to the length of the screen’s diagonal.) HINT: Use the Pythagorean Theorem. Round your answers to the nearest hundredth.
  8. Real Life Application When an object is dropped, its speed continually increases until it reaches the ground. This scenario can be modeled by the equation h=-16t^2+h_0 , where h is the height, t is the time (in seconds), and h_0 is the initial height of the object. Round your answers to the nearest hundredth.
    1. If you drop a ball from 200 feet, what is the height after 2 seconds?
    2. After how many seconds will the ball hit the ground?

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