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2.5 Quadratic equations  (Page 5/14)

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Solving a quadratic equation with the quadratic formula

Use the quadratic formula to solve   x 2 + x + 2 = 0.

First, we identify the coefficients:   a = 1 , b = 1 , and   c = 2.

Substitute these values into the quadratic formula.

x = − b ± b 2 − 4 a c 2 a = − ( 1 ) ± ( 1 ) 2 − ( 4 ) ⋅ ( 1 ) ⋅ ( 2 ) 2 ⋅ 1 = − 1 ± 1 − 8 2 = − 1 ± − 7 2 = − 1 ± i 7 2

The solutions to the equation are   − 1 + i 7 2   and   − 1 − i 7 2   or   − 1 2 + i 7 2   and   − 1 2 − i 7 2 .

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Solve the quadratic equation using the quadratic formula:   9 x 2 + 3 x − 2 = 0.

x = − 2 3 , x = 1 3

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The discriminant

The quadratic formula    not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant    , or the expression under the radical,   b 2 − 4 a c .   The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. [link] relates the value of the discriminant to the solutions of a quadratic equation.

Value of Discriminant Results
b 2 − 4 a c = 0 One rational solution (double solution)
b 2 − 4 a c > 0 , perfect square Two rational solutions
b 2 − 4 a c > 0 , not a perfect square Two irrational solutions
b 2 − 4 a c < 0 Two complex solutions

The discriminant

For   a x 2 + b x + c = 0 , where   a , b , and   c   are rational and real numbers, the discriminant    is the expression under the radical in the quadratic formula:   b 2 − 4 a c .   It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Using the discriminant to find the nature of the solutions to a quadratic equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

  1. x 2 + 4 x + 4 = 0
  2. 8 x 2 + 14 x + 3 = 0
  3. 3 x 2 − 5 x − 2 = 0
  4. 3 x 2 − 10 x + 15 = 0

Calculate the discriminant   b 2 − 4 a c   for each equation and state the expected type of solutions.

  1. x 2 + 4 x + 4 = 0

    b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0.   There will be one rational double solution.

  2. 8 x 2 + 14 x + 3 = 0

    b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100.   As   100   is a perfect square, there will be two rational solutions.

  3. 3 x 2 − 5 x − 2 = 0

    b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49.   As   49   is a perfect square, there will be two rational solutions.

  4. 3 x 2 −10 x + 15 = 0

    b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80.   There will be two complex solutions.

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Using the pythagorean theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem    . It is based on a right triangle, and states the relationship among the lengths of the sides as   a 2 + b 2 = c 2 , where   a   and   b   refer to the legs of a right triangle adjacent to the   90°   angle, and   c   refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

a 2 + b 2 = c 2

where   a   and   b   refer to the legs of a right triangle adjacent to the   90 ∘   angle, and   c   refers to the hypotenuse, as shown in [link] .

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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