Math Problem Statement

find the roots of the quadratic equation by completing the square

Solution

To solve these quadratic equations by completing the square, we'll go through each step by step. Here are the quadratic equations:

  1. x2+6x+5=0x^2 + 6x + 5 = 0
  2. x2+4x6=0x^2 + 4x - 6 = 0
  3. x28x20=0x^2 - 8x - 20 = 0
  4. x26x10=0x^2 - 6x - 10 = 0

1. x2+6x+5=0x^2 + 6x + 5 = 0

  • Move the constant term to the other side: x2+6x=5x^2 + 6x = -5
  • Take half of the coefficient of xx, square it, and add it to both sides: (62)2=9\left(\frac{6}{2}\right)^2 = 9 x2+6x+9=5+9x^2 + 6x + 9 = -5 + 9
  • Rewrite the left side as a perfect square: (x+3)2=4(x + 3)^2 = 4
  • Take the square root of both sides: x+3=±2x + 3 = \pm 2
  • Solve for xx: x=3±2x = -3 \pm 2 So, x=1x = -1 or x=5x = -5.

2. x2+4x6=0x^2 + 4x - 6 = 0

  • Move the constant term to the other side: x2+4x=6x^2 + 4x = 6
  • Take half of the coefficient of xx, square it, and add it to both sides: (42)2=4\left(\frac{4}{2}\right)^2 = 4 x2+4x+4=6+4x^2 + 4x + 4 = 6 + 4
  • Rewrite the left side as a perfect square: (x+2)2=10(x + 2)^2 = 10
  • Take the square root of both sides: x+2=±10x + 2 = \pm \sqrt{10}
  • Solve for xx: x=2±10x = -2 \pm \sqrt{10} So, x=2+10x = -2 + \sqrt{10} or x=210x = -2 - \sqrt{10}.

3. x28x20=0x^2 - 8x - 20 = 0

  • Move the constant term to the other side: x28x=20x^2 - 8x = 20
  • Take half of the coefficient of xx, square it, and add it to both sides: (82)2=16\left(\frac{-8}{2}\right)^2 = 16 x28x+16=20+16x^2 - 8x + 16 = 20 + 16
  • Rewrite the left side as a perfect square: (x4)2=36(x - 4)^2 = 36
  • Take the square root of both sides: x4=±6x - 4 = \pm 6
  • Solve for xx: x=4±6x = 4 \pm 6 So, x=10x = 10 or x=2x = -2.

4. x26x10=0x^2 - 6x - 10 = 0

  • Move the constant term to the other side: x26x=10x^2 - 6x = 10
  • Take half of the coefficient of xx, square it, and add it to both sides: (62)2=9\left(\frac{-6}{2}\right)^2 = 9 x26x+9=10+9x^2 - 6x + 9 = 10 + 9
  • Rewrite the left side as a perfect square: (x3)2=19(x - 3)^2 = 19
  • Take the square root of both sides: x3=±19x - 3 = \pm \sqrt{19}
  • Solve for xx: x=3±19x = 3 \pm \sqrt{19} So, x=3+19x = 3 + \sqrt{19} or x=319x = 3 - \sqrt{19}.

Summary of Solutions:

  1. x=1x = -1 or x=5x = -5
  2. x=2+10x = -2 + \sqrt{10} or x=210x = -2 - \sqrt{10}
  3. x=10x = 10 or x=2x = -2
  4. x=3+19x = 3 + \sqrt{19} or x=319x = 3 - \sqrt{19}

Would you like further details on any of these solutions or have any other questions?

Here are five related questions for further practice:

  1. What is the general formula for completing the square for any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0?
  2. Can you apply completing the square to solve the quadratic equation 2x2+8x10=02x^2 + 8x - 10 = 0?
  3. How does the method of completing the square relate to the quadratic formula?
  4. What are the conditions under which completing the square might be more advantageous than using the quadratic formula?
  5. How can completing the square be used to derive the vertex form of a quadratic function?

Tip: Completing the square is also useful for converting quadratic functions into vertex form, which makes it easier to find the vertex and analyze the graph of the function.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Completing the Square
Roots of Equations

Formulas

Quadratic equation completing the square

Theorems

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Suitable Grade Level

Grades 9-12