Solving Quadratic Equations Using the Quadratic Formula

Learning Objective

Introduction

We can solve any quadratic equation by completing the squareThe process of changing a polynomial of the form `x^2+bx` into a perfect square trinomial `x^2+bx+(b/2)^2`, or `(x+b/2)^2`.—turning a polynomial into a perfect square trinomial. If we complete the square on the generic equation `ax^2+bx+c=0` and then solve for `x`, we find that `x=(-b+-sqrt(b^2-4ac))/(2a)`. This rather awkward-looking equation is known as the quadratic formulaThe formula `x=(-b+-sqrt(b^2-4ac))/(2a)`; it is used to solve a quadratic equation of the form `ax^2+bx+c=0`..

This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The quadratic formula can be used to solve any quadratic equation of the form `ax^2+bx+c=0`

Deriving the Quadratic Formula

Let's complete the square on the general equation, `ax^2+bx+c=0`, and see exactly how that produces the quadratic formula. Recall the process of completing the square:

Can you complete the square on the general quadratic equation `ax^2+bx+c=0`? Try it yourself before you continue to the example below. Hint: When working with the general equation `ax^2+bx+c=0`, there is an added complication, in that the coefficient of `x^2` is not equal to `1`. You can divide the equation by `a`, which makes some of the expressions a bit messy, but if you are careful, everything will work out, and at the end, you’ll have the quadratic formula!

Example

Problem

Complete the square of `ax^2+bx+c=0` to arrive at the quadratic formula.

 

`x^2+b/ax+c/a=0`

 

Divide both sides of the equation by `a`, so that the coefficient of `x^2` is `1`.

 

`x^2+b/ax=-c/a`

 

 

Rewrite so the left side is in the form `x^2+bx` (although in this case `bx` is `b/ax`).

 

`x^2+b/ax+(b/(2a))^2=-c/a+(b/(2a))^2`

 

 

Add `(b/(2a))^2` to both sides to complete the square.

 

`(x+b/(2a))^2=-c/a+(b/(2a))^2`

 

Write the left side as a binomial squared.

 

`(x+b/(2a))^2=-c/a+(b^2)/(4a^2)`

 

Evaluate `(b/(2a))^2` as `b^2/(4a^2)`

 

`(x+b/(2a))^2=-(4ac)/(4a^2)+(b^2)/(4a^2)`

 

Write the fractions on the right side using a common denominator.

 

`(x+b/(2a))^2=(b^2-4ac)/(4a^2)`

 

Add the fractions on the right.

 

`x+b/(2a)=+-sqrt((b^2-4ac)/(4a^2))`

 

Take the square root of both sides. Remember that you want both the positive and negative square roots!

 

`x=-b/(2a)+-sqrt((b^2-4ac)/(4a^2))`

 

Subtract `b/(2a)` from both sides to isolate `x`.

 

`x=-b/(2a)+-(sqrt(b^2-4ac))/(2a)`

 

The denominator under the radical is a perfect square, so: `sqrt((b^2-4ac)/(4a^2))=sqrt(b^2-4ac)/sqrt(4a^2)=sqrt(b^2-4ac)/(2a)` 

 

`x=(-b+-sqrt(b^2-4ac))/(2a)`

 

Add the fractions since they have a common denominator.

Answer

`x=(-b+-sqrt(b^2-4ac))/(2a)`

 

 

           

There we have it, the quadratic formula.

Solving a Quadratic Equation using the Quadratic Formula

The quadratic formula will work with any quadratic equation, but only if the equation is in standard form, `ax^2+bx+c=0`. To use it, follow these steps:

That's a lot of steps. Let’s try them:

Example

Problem

Use the quadratic formula to solve the equation `3x^2-11x-4=0`.  

 

`3x^2` `-11x` `-4` `=` `0`
Downward facing arrow connecting 3 X squared above to A X squared below. Downward facing arrow connecting minus 11 X above to plus B X below. Downward facing arrow connecting minus 4 above to plus C below.    
`ax^2` `+bx` `+c` `=` `0`

 

 

 

 

`a=3`, `b=-11``c=-4`

 

Note that the subtraction signs mean the coefficients `b` and `c` are negative.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-11)+-sqrt((-11)^2-4(3)(-4)))/(2(3))`

 

Substitute values into the quadratic formula.

`x=(11+-sqrt(121+48))/6`

 

Simplify, being careful to get the signs correct.

`x=(11+-sqrt169)/6`

 

Simplify some more.

`x=(11+-13)/6`

 

Simplify the radical: `sqrt169=13`

`x=(11+13)/6=24/6=4`

or

`x=(11-13)/6=(-2)/6=-1/3`

 

Separate and simplify to find the solutions to the quadratic equation. Note that in one, `13` is added and in the other, `13` is subtracted.

Answer

`x = 4` or `-1/3` 

 

 

         

The solution to the quadratic equation gives the `x`-coordinates of the x-interceptsThe point where a line meets or crosses the `x`-axis., or the roots of a quadratic equationThe `x`-intercepts of the parabola or the solution of the equation.. The roots of the quadratic equation are the values where the parabola crosses the `x`-axis. We can check this by looking at the graph of the function `y=3x^2-11x-4` and see that the roots are `(4, 0)` and `(-1/3,0)`

On a coordinate plane, a parabola opens upward. It intersects the x-axis at two points, (negative zero point 33, 0) and (4, 0),  and is labeled with the formula, “3 x squared minus 11 x minus 4.” The vertex is near (negative 14, 2).

The example above shows a quadratic equation with two solutions. Below is an example with one solution. Compare the simplified radicals of the two examples:

Example

Problem

Use the quadratic formula to solve the equation `x^2-2x=6x-16`.  

 

 

`x^2-8x+16=0`

 

 

Subtract `6x` from each side and add `16` to both sides to put the equation in the form `ax^2+bx+c=0`.

 

`x^2` `-8x` `+16` `=` `0`
Downward facing arrow connecting X squared above to A X squared below. Downward facing arrow connecting minus 8 X above to plus B X below. Downward facing arrow connecting plus 16 above to plus C below.    
`ax^2` `+bx` `+c` `=` `0`

 

 

 

 

Identify the coefficients `a``b`, and `c``x^2 = 1x^2`, so `a = 1`. Since `8x` is subtracted, `b` is negative.

`a=1`, `b=-8``c=16`

 

`x=(-(-8)+-sqrt((-8)^2-4(1)(16)))/(2(1))`

 

Apply the quadratic formula.

 

`x=(8+-sqrt(64-64))/2`

 

Simplify.

 

`x=(8+-sqrt0)/2=8/2=4`

 

Since the square root of `0` is `0`, and adding or subtracting `0` both give the same result, there is only one possible value.

Answer

`x = 4`

 

 

           

This quadratic equation has only one solution, so the graph of the function `y=x^2-8x+16` will touch the `x`-axis once. It has only one root.

On a coordinate plane, a parabola opens upward. It has a vertex on the x-axis at (4, 0) and is labeled with the formula, “x squared minus 8 x plus 16.”

Something to notice—the quadratic equation `x^2-8x+16=0` can be factored to `(x-4)^2=0`. So although the quadratic formula gave us the solution, it would have been faster to factor it. It's worth checking to see if a quadratic equation can be easily factored before applying the quadratic formula.

Use the quadratic formula to solve the equation `x^2-2x-4=0`.  

 

A) `x = 2`

 

B) `x=11``x = -9`

 

C) `x=2+sqrt5``x=2-sqrt5` 

 

D) `x=1+sqrt5``x=1-sqrt5` 

 

The Discriminant

A quadratic equation may have two roots, one root, or no roots. In the quadratic formula, the expression underneath the radical symbol determines how many solutions the formula will have. This expression, `b^2-4ac`, is called the discriminantThe expression `b^2 - 4ac` under the radical in the quadratic formula; the expression can be used to determine the number of real roots the quadratic equation has. of the equation `ax^2+bx+c=0`.  

Let’s think about how `b^2-4ac` will affect the evaluation of `sqrt(b^2-4ac)`, and how it helps to determine the solution set.

Example

Problem

Use the discriminant to determine whether the quadratic equation `x^2-4x+10=0` has two, one, or no solutions.

 

`(-4)^2-4(1)(10)`

 

 

 

Evaluate `b^2-4ac`.  

`a = 1`, `b = -4`, and `c = 10`

 

`16-40=-24`

 

 

 

 

The result is a negative number. The discriminant is negative, so the quadratic equation has no solution.

Answer

no solution

 

 

         

Suppose a quadratic equation has a discriminant that evaluates to zero. Which of the following statements is always true?

 

A) The equation has two solutions and the graph of the parabola will open upward.

 

B) The equation has one solution and the graph of the parabola will open downward.

 

C) The equation has one solution and you can’t tell anything about the direction of the parabola.

 

D) The equation has zero solutions and you can’t tell anything about the direction of the parabola.

 

 

Summary

The quadratic formula, `x=(-b+-sqrt(b^2-4ac))/(2a)`, is found by completing the square of the quadratic equation `ax^2+bx+c=0`. The formula can be used to find the solution to a quadratic equation and to identify any possible roots, or `x`-intercepts, of the function.

The discriminant of the quadratic formula is the quantity under the radical, `b^2-4ac`. It determines how many solutions there are to the quadratic equation. If the discriminant is positive, there are `2` roots. If it is zero, there is `1` root. If the discriminant is negative, there are `0` roots.