Learn on PengienVision, Algebra 1Chapter 9: Solving Quadratic Equations

Lesson 4: Solving Quadratic Equations Using Square Roots

Grade 11 students in enVision Algebra 1 learn how to solve quadratic equations using square roots, covering equation forms such as x² = a, ax² = c, and ax² + b = c. The lesson teaches students to isolate the variable and apply both positive and negative square roots, including recognizing when no real solution exists due to a negative radicand. Real-world applications, such as using the Pythagorean theorem to find the height of a cell phone tower, reinforce how this method is used to solve practical problems.

Section 1

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 2

Solving equations of form ax2=kax^2 = k

Property

To solve a quadratic equation using the square root property:
Step 1. Isolate the quadratic term and make its coefficient one. For an equation like ax2=kax^2 = k, this means rewriting it as x2=kax^2 = \frac{k}{a}.
Step 2. Use the Square Root Property: x=±kax = \pm\sqrt{\frac{k}{a}}.
Step 3. Simplify the radical.
Step 4. Check the solutions.

Examples

  • Solve 5x2=1255x^2 = 125. First, divide by 5 to isolate x2x^2, giving x2=25x^2 = 25. Then, use the Square Root Property to find x=±25x = \pm\sqrt{25}, so the solutions are x=5x=5 and x=5x=-5.
  • Solve 34u2+2=29\frac{3}{4}u^2 + 2 = 29. Subtract 2 to get 34u2=27\frac{3}{4}u^2 = 27. Multiply by 43\frac{4}{3} to get u2=36u^2 = 36. The solutions are u=±36u = \pm\sqrt{36}, so u=6u=6 and u=6u=-6.
  • Solve 3x29=543x^2 - 9 = 54. Add 9 to get 3x2=633x^2 = 63. Divide by 3 to get x2=21x^2 = 21. The solutions are x=±21x = \pm\sqrt{21}.

Explanation

Before you can use the Square Root Property, the x2x^2 term must be by itself. This method is about clearing away any coefficients or constants first. Once you have x2x^2 isolated, you can solve for the two possible values of xx.

Section 3

Complete Step-by-Step Solution Process for ax² + b = c

Property

To solve equations of the form ax2+b=cax² + b = c, follow these steps:

  1. Subtract bb from both sides: ax2=cbax² = c - b
  2. Divide both sides by aa: x2=cbax² = \frac{c - b}{a}
  3. Take the square root of both sides: x=±cbax = ±\sqrt{\frac{c - b}{a}} (when cba0\frac{c - b}{a} \geq 0)

Examples

Book overview

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Chapter 9: Solving Quadratic Equations

  1. Lesson 1

    Lesson 1: Solving Quadratic Equations Using Graphs and Tables

  2. Lesson 2

    Lesson 2: Solving Quadratic Equations by Factoring

  3. Lesson 3

    Lesson 3: Rewriting Radical Expressions

  4. Lesson 4Current

    Lesson 4: Solving Quadratic Equations Using Square Roots

  5. Lesson 5

    Lesson 5: Completing the Square

  6. Lesson 6

    Lesson 6: The Quadratic Formula and the Discriminant

  7. Lesson 7

    Lesson 7: Solving Nonlinear Systems of Equations

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

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 2

Solving equations of form ax2=kax^2 = k

Property

To solve a quadratic equation using the square root property:
Step 1. Isolate the quadratic term and make its coefficient one. For an equation like ax2=kax^2 = k, this means rewriting it as x2=kax^2 = \frac{k}{a}.
Step 2. Use the Square Root Property: x=±kax = \pm\sqrt{\frac{k}{a}}.
Step 3. Simplify the radical.
Step 4. Check the solutions.

Examples

  • Solve 5x2=1255x^2 = 125. First, divide by 5 to isolate x2x^2, giving x2=25x^2 = 25. Then, use the Square Root Property to find x=±25x = \pm\sqrt{25}, so the solutions are x=5x=5 and x=5x=-5.
  • Solve 34u2+2=29\frac{3}{4}u^2 + 2 = 29. Subtract 2 to get 34u2=27\frac{3}{4}u^2 = 27. Multiply by 43\frac{4}{3} to get u2=36u^2 = 36. The solutions are u=±36u = \pm\sqrt{36}, so u=6u=6 and u=6u=-6.
  • Solve 3x29=543x^2 - 9 = 54. Add 9 to get 3x2=633x^2 = 63. Divide by 3 to get x2=21x^2 = 21. The solutions are x=±21x = \pm\sqrt{21}.

Explanation

Before you can use the Square Root Property, the x2x^2 term must be by itself. This method is about clearing away any coefficients or constants first. Once you have x2x^2 isolated, you can solve for the two possible values of xx.

Section 3

Complete Step-by-Step Solution Process for ax² + b = c

Property

To solve equations of the form ax2+b=cax² + b = c, follow these steps:

  1. Subtract bb from both sides: ax2=cbax² = c - b
  2. Divide both sides by aa: x2=cbax² = \frac{c - b}{a}
  3. Take the square root of both sides: x=±cbax = ±\sqrt{\frac{c - b}{a}} (when cba0\frac{c - b}{a} \geq 0)

Examples

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: Solving Quadratic Equations Using Graphs and Tables

  2. Lesson 2

    Lesson 2: Solving Quadratic Equations by Factoring

  3. Lesson 3

    Lesson 3: Rewriting Radical Expressions

  4. Lesson 4Current

    Lesson 4: Solving Quadratic Equations Using Square Roots

  5. Lesson 5

    Lesson 5: Completing the Square

  6. Lesson 6

    Lesson 6: The Quadratic Formula and the Discriminant

  7. Lesson 7

    Lesson 7: Solving Nonlinear Systems of Equations