Math Problem Statement

Solve the following quadratic equations using the quadratic formula.

Solution

The image contains six quadratic equations that need to be solved using the quadratic formula. Here's a step-by-step breakdown of each equation:

Quadratic Formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Where aa, bb, and cc are coefficients from the general quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.

Solutions:

  1. Equation (a): x2+x1=0x^2 + x - 1 = 0

    • Coefficients: a=1a = 1, b=1b = 1, c=1c = -1
    • Using the quadratic formula: x=1±124(1)(1)2(1)=1±1+42=1±52x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} Solutions: x=1+52,x=152x = \frac{-1 + \sqrt{5}}{2}, \quad x = \frac{-1 - \sqrt{5}}{2}
  2. Equation (b): 2x2=5x+32x^2 = 5x + 3

    Rearranging: 2x25x3=02x^2 - 5x - 3 = 0

    • Coefficients: a=2a = 2, b=5b = -5, c=3c = -3
    • Applying the quadratic formula: x=(5)±(5)24(2)(3)2(2)=5±25+244=5±494x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} Solutions: x=5+74=3,x=574=12x = \frac{5 + 7}{4} = 3, \quad x = \frac{5 - 7}{4} = -\frac{1}{2}
  3. Equation (c): 2x2+x5=02x^2 + x - 5 = 0

    • Coefficients: a=2a = 2, b=1b = 1, c=5c = -5
    • Applying the quadratic formula: x=1±124(2)(5)2(2)=1±1+404=1±414x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-5)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 40}}{4} = \frac{-1 \pm \sqrt{41}}{4} Solutions: x=1+414,x=1414x = \frac{-1 + \sqrt{41}}{4}, \quad x = \frac{-1 - \sqrt{41}}{4}
  4. Equation (d): x2+7x1=0x^2 + 7x - 1 = 0

    • Coefficients: a=1a = 1, b=7b = 7, c=1c = -1
    • Applying the quadratic formula: x=7±724(1)(1)2(1)=7±49+42=7±532x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-1)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 4}}{2} = \frac{-7 \pm \sqrt{53}}{2} Solutions: x=7+532,x=7532x = \frac{-7 + \sqrt{53}}{2}, \quad x = \frac{-7 - \sqrt{53}}{2}
  5. Equation (e): (x+1)(x5)=3(x+1)(x-5) = -3

    Expanding: x25x+x5=3x24x5=3x^2 - 5x + x - 5 = -3 \quad \Rightarrow \quad x^2 - 4x - 5 = -3 Rearranging: x24x2=0x^2 - 4x - 2 = 0

    • Coefficients: a=1a = 1, b=4b = -4, c=2c = -2
    • Applying the quadratic formula: x=(4)±(4)24(1)(2)2(1)=4±16+82=4±242=4±262x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)} = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} Solutions: x=2+6,x=26x = 2 + \sqrt{6}, \quad x = 2 - \sqrt{6}
  6. Equation (f): 3x2=4x3x^2 = 4x

    Rearranging: 3x24x=0x(3x4)=03x^2 - 4x = 0 \quad \Rightarrow \quad x(3x - 4) = 0 Solutions: x=0,3x4=0x=43x = 0, \quad 3x - 4 = 0 \quad \Rightarrow \quad x = \frac{4}{3}


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Related Questions:

  1. How do you derive the quadratic formula?
  2. What are the discriminant's roles in quadratic equations?
  3. How do you solve quadratic equations by factoring?
  4. Can the quadratic formula give complex solutions?
  5. When is completing the square useful in solving quadratic equations?

Tip:

Always check the discriminant (b24acb^2 - 4ac) before solving to determine the nature of the roots (real or complex).

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Algebra

Formulas

Quadratic Formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-11