Learn on PengienVision, Algebra 2Chapter 2: Quadratic Functions and Equations

Lesson 6: The Quadratic Formula

In this Grade 11 enVision Algebra 2 lesson, students learn to solve quadratic equations using the Quadratic Formula, x = (−b ± √(b²−4ac)) / 2a, including cases that produce real and complex (imaginary) solutions. The lesson also introduces the discriminant, b²−4ac, as a tool for predicting the number and type of roots without fully solving the equation. Students practice applying the formula and compare it to factoring to determine the most efficient solution method for a given equation.

Section 1

Deriving the Quadratic Formula

Property

The quadratic formula x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} can be derived by completing the square on the general quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.

Examples

Section 2

Choosing a solution method

Property

  1. Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
  2. Step 2. Try the Square Root Property next. If the equation fits the form ax2=kax^2 = k or a(x−h)2=ka(x-h)^2 = k, it can easily be solved by using the Square Root Property.
  3. Step 3. Use the Quadratic Formula. Any quadratic equation can be solved by using the Quadratic Formula.

Examples

  • For x2−9x+20=0x^2 - 9x + 20 = 0, Factoring is best. The equation factors into (x−4)(x−5)=0(x-4)(x-5)=0, giving solutions x=4x=4 and x=5x=5 quickly.
  • For 2(x−3)2=322(x-3)^2 = 32, the Square Root Property is ideal. Divide by 2 to get (x−3)2=16(x-3)^2=16, then take the square root: x−3=±4x-3 = \pm 4, so x=7x=7 or x=−1x=-1.
  • For 5x2+7x−11=05x^2 + 7x - 11 = 0, the numbers do not allow for easy factoring. The Quadratic Formula is the most appropriate and reliable method to find the solutions.

Explanation

While the Quadratic Formula is a universal solver, it is not always the fastest. Always check if an equation can be factored or if the Square Root Property applies. Choosing the right tool makes solving equations much simpler.

Section 3

Using the discriminant

Property

In the Quadratic Formula, x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, the quantity b2−4acb^2 - 4ac is called the discriminant.
For a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, a≠0a \neq 0:

  • If b2−4ac>0b^2 - 4ac > 0, the equation has 2 real solutions.
  • If b2−4ac=0b^2 - 4ac = 0, the equation has 1 real solution.
  • If b2−4ac<0b^2 - 4ac < 0, the equation has 2 complex solutions.

Examples

  • For the equation 3x2+5x−8=03x^2 + 5x - 8 = 0, the discriminant is (5)2−4(3)(−8)=25+96=121(5)^2 - 4(3)(-8) = 25 + 96 = 121. Since 121>0121 > 0, there are 2 real solutions.
  • For the equation 4y2−28y+49=04y^2 - 28y + 49 = 0, the discriminant is (−28)2−4(4)(49)=784−784=0(-28)^2 - 4(4)(49) = 784 - 784 = 0. Since the discriminant is 0, there is 1 real solution.
  • For the equation 2n2+3n+5=02n^2 + 3n + 5 = 0, the discriminant is (3)2−4(2)(5)=9−40=−31(3)^2 - 4(2)(5) = 9 - 40 = -31. Since −31<0-31 < 0, there are 2 complex solutions.

Explanation

The discriminant is the part of the Quadratic Formula under the radical. Its value predicts the number and type of solutions without having to solve the entire equation. A positive, zero, or negative result tells you exactly what kind of answers to expect.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Deriving the Quadratic Formula

Property

The quadratic formula x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} can be derived by completing the square on the general quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.

Examples

Section 2

Choosing a solution method

Property

  1. Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
  2. Step 2. Try the Square Root Property next. If the equation fits the form ax2=kax^2 = k or a(x−h)2=ka(x-h)^2 = k, it can easily be solved by using the Square Root Property.
  3. Step 3. Use the Quadratic Formula. Any quadratic equation can be solved by using the Quadratic Formula.

Examples

  • For x2−9x+20=0x^2 - 9x + 20 = 0, Factoring is best. The equation factors into (x−4)(x−5)=0(x-4)(x-5)=0, giving solutions x=4x=4 and x=5x=5 quickly.
  • For 2(x−3)2=322(x-3)^2 = 32, the Square Root Property is ideal. Divide by 2 to get (x−3)2=16(x-3)^2=16, then take the square root: x−3=±4x-3 = \pm 4, so x=7x=7 or x=−1x=-1.
  • For 5x2+7x−11=05x^2 + 7x - 11 = 0, the numbers do not allow for easy factoring. The Quadratic Formula is the most appropriate and reliable method to find the solutions.

Explanation

While the Quadratic Formula is a universal solver, it is not always the fastest. Always check if an equation can be factored or if the Square Root Property applies. Choosing the right tool makes solving equations much simpler.

Section 3

Using the discriminant

Property

In the Quadratic Formula, x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, the quantity b2−4acb^2 - 4ac is called the discriminant.
For a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, a≠0a \neq 0:

  • If b2−4ac>0b^2 - 4ac > 0, the equation has 2 real solutions.
  • If b2−4ac=0b^2 - 4ac = 0, the equation has 1 real solution.
  • If b2−4ac<0b^2 - 4ac < 0, the equation has 2 complex solutions.

Examples

  • For the equation 3x2+5x−8=03x^2 + 5x - 8 = 0, the discriminant is (5)2−4(3)(−8)=25+96=121(5)^2 - 4(3)(-8) = 25 + 96 = 121. Since 121>0121 > 0, there are 2 real solutions.
  • For the equation 4y2−28y+49=04y^2 - 28y + 49 = 0, the discriminant is (−28)2−4(4)(49)=784−784=0(-28)^2 - 4(4)(49) = 784 - 784 = 0. Since the discriminant is 0, there is 1 real solution.
  • For the equation 2n2+3n+5=02n^2 + 3n + 5 = 0, the discriminant is (3)2−4(2)(5)=9−40=−31(3)^2 - 4(2)(5) = 9 - 40 = -31. Since −31<0-31 < 0, there are 2 complex solutions.

Explanation

The discriminant is the part of the Quadratic Formula under the radical. Its value predicts the number and type of solutions without having to solve the entire equation. A positive, zero, or negative result tells you exactly what kind of answers to expect.