Learn on PengiBig Ideas Math, Algebra 1Chapter 9: Solving Quadratic Equations

Lesson 3: Solving Quadratic Equations Using Square Roots

Property If $x^2 = k$, and $k \ge 0$, then $x = \sqrt{k}$ or $x = \sqrt{k}$. The solution can also be written as $x = \pm \sqrt{k}$.

Section 1

Square Root Property

Property

If x2=kx^2 = k, and k0k \ge 0, then x=kx = \sqrt{k} or x=kx = -\sqrt{k}. The solution can also be written as x=±kx = \pm \sqrt{k}.

Examples

  • To solve x2=81x^2 = 81, we apply the Square Root Property to get x=±81x = \pm\sqrt{81}, which gives the solutions x=9x = 9 and x=9x = -9.
  • For the equation y2=15y^2 = 15, the solutions are y=±15y = \pm\sqrt{15}, which we leave in radical form as y=15y = \sqrt{15} and y=15y = -\sqrt{15}.
  • If z2=9z^2 = -9, there is no real solution because the square root of a negative number is not a real number.

Explanation

When a variable squared equals a number, the variable itself can be either the positive or negative square root of that number. This is because squaring a negative value results in a positive value, creating two possible answers.

Section 2

Solving Equations of the Form x^2 = p

Property

Taking a square root is the opposite of squaring a number.
To solve an equation of the form x2=kx^2 = k (where k>0k > 0), we take the square root of both sides.
Because a positive number has two square roots, the solution is written as:

x=±kx = \pm\sqrt{k}

Examples

  • To solve the equation x2=81x^2 = 81, we take the square root of both sides. The solutions are x=±81x = \pm\sqrt{81}, which means x=9x = 9 and x=9x = -9.

Section 3

Solve equations of the form ax2=kax^2 = k

Property

To solve a quadratic equation using the Square Root Property:

  1. Isolate the quadratic term and make its coefficient one.
  2. Use the Square Root Property.
  3. Simplify the radical.
  4. Check the solutions.

Examples

  • To solve 4x2=1004x^2 = 100, first divide by 4 to get x2=25x^2 = 25. Then, using the Square Root Property, x=±25x = \pm\sqrt{25}, so x=5x = 5 and x=5x = -5.
  • Solve 3y254=03y^2 - 54 = 0. First, add 54 to both sides to get 3y2=543y^2 = 54. Divide by 3 to get y2=18y^2 = 18. The solution is y=±18=±32y = \pm\sqrt{18} = \pm 3\sqrt{2}.
  • Solve 3c2=753c^2 = 75. Divide by 3 to get c2=25c^2=25. Using the property, c=±25c = \pm\sqrt{25}, so c=5,c=5c=5, c=-5.

Explanation

Before you can apply the Square Root Property, the x2x^2 term must be by itself. To achieve this, divide both sides of the equation by the coefficient of x2x^2, and then proceed with taking the square root.

Book overview

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Chapter 9: Solving Quadratic Equations

  1. Lesson 1

    Lesson 1: Properties of Radicals

  2. Lesson 2

    Lesson 2: Solving Quadratic Equations by Graphing

  3. Lesson 3Current

    Lesson 3: Solving Quadratic Equations Using Square Roots

  4. Lesson 4

    Lesson 4: Solving Quadratic Equations by Completing the Square

  5. Lesson 5

    Lesson 5: Solving Quadratic Equations Using the Quadratic Formula

  6. Lesson 6

    Lesson 6: Solving Nonlinear Systems of Equations

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

Square Root Property

Property

If x2=kx^2 = k, and k0k \ge 0, then x=kx = \sqrt{k} or x=kx = -\sqrt{k}. The solution can also be written as x=±kx = \pm \sqrt{k}.

Examples

  • To solve x2=81x^2 = 81, we apply the Square Root Property to get x=±81x = \pm\sqrt{81}, which gives the solutions x=9x = 9 and x=9x = -9.
  • For the equation y2=15y^2 = 15, the solutions are y=±15y = \pm\sqrt{15}, which we leave in radical form as y=15y = \sqrt{15} and y=15y = -\sqrt{15}.
  • If z2=9z^2 = -9, there is no real solution because the square root of a negative number is not a real number.

Explanation

When a variable squared equals a number, the variable itself can be either the positive or negative square root of that number. This is because squaring a negative value results in a positive value, creating two possible answers.

Section 2

Solving Equations of the Form x^2 = p

Property

Taking a square root is the opposite of squaring a number.
To solve an equation of the form x2=kx^2 = k (where k>0k > 0), we take the square root of both sides.
Because a positive number has two square roots, the solution is written as:

x=±kx = \pm\sqrt{k}

Examples

  • To solve the equation x2=81x^2 = 81, we take the square root of both sides. The solutions are x=±81x = \pm\sqrt{81}, which means x=9x = 9 and x=9x = -9.

Section 3

Solve equations of the form ax2=kax^2 = k

Property

To solve a quadratic equation using the Square Root Property:

  1. Isolate the quadratic term and make its coefficient one.
  2. Use the Square Root Property.
  3. Simplify the radical.
  4. Check the solutions.

Examples

  • To solve 4x2=1004x^2 = 100, first divide by 4 to get x2=25x^2 = 25. Then, using the Square Root Property, x=±25x = \pm\sqrt{25}, so x=5x = 5 and x=5x = -5.
  • Solve 3y254=03y^2 - 54 = 0. First, add 54 to both sides to get 3y2=543y^2 = 54. Divide by 3 to get y2=18y^2 = 18. The solution is y=±18=±32y = \pm\sqrt{18} = \pm 3\sqrt{2}.
  • Solve 3c2=753c^2 = 75. Divide by 3 to get c2=25c^2=25. Using the property, c=±25c = \pm\sqrt{25}, so c=5,c=5c=5, c=-5.

Explanation

Before you can apply the Square Root Property, the x2x^2 term must be by itself. To achieve this, divide both sides of the equation by the coefficient of x2x^2, and then proceed with taking the square root.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 9: Solving Quadratic Equations

  1. Lesson 1

    Lesson 1: Properties of Radicals

  2. Lesson 2

    Lesson 2: Solving Quadratic Equations by Graphing

  3. Lesson 3Current

    Lesson 3: Solving Quadratic Equations Using Square Roots

  4. Lesson 4

    Lesson 4: Solving Quadratic Equations by Completing the Square

  5. Lesson 5

    Lesson 5: Solving Quadratic Equations Using the Quadratic Formula

  6. Lesson 6

    Lesson 6: Solving Nonlinear Systems of Equations