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1.05 Quadratic formula

Lesson

There are four ways to solve a quadratic equation (i.e. an equation of the form $ax^2+bx+c=0$ax2+bx+c=0): 

  • By algebraic manipulation
  • By factoring and using the null factor law
  • By completing the square
  • By using the quadratic formula
     
Quadratic formula

If $ax^2+bx+c=0$ax2+bx+c=0, then:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=b±b24ac2a

The advantage of using the quadratic formula is that it always works (unlike factoring) and it always follows the exact same process. However, the other methods can be more efficient in many cases. 

The quadratic formula might seem quite complex when you first come across it, but it can be broken down into smaller parts.

  • The ± allows for the possibility of two solutions.
  • The $b^2-4ac$b24ac under the square root sign is important as it will tell us how many solutions there are. This is known as the discriminant.

 

Practice questions

Question 1

Solve the equation $x^2-5x+6=0$x25x+6=0 by using the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=b±b24ac2a.

Write each solution on the same line, separated by a comma.

Question 2

Solve the equation $25x^2+30x=-9$25x2+30x=9 using the quadratic formula.

 

Properties of the discriminant

As we have seen with quadratic equations of the form  $ax^2+bx+c=0$ax2+bx+c=0 there can be two, one or no real solutions. If we think about the graphs of quadratics, this means that there can be two, one or no $x$x-intercepts. This is because the solutions to a function in the form $y=ax^2+bx+c$y=ax2+bx+c are the places where the function crosses the $x$x-axis, hence the $y$y value is $0$0.

No solutions, one solution and two solutions

Looking at the image above, we can see that a quadratic equation can have either:

  • Two real solutions, these are the two zeros or $x$x-intercepts, where the graph of the quadratic function passes through the $x$x-axis.
  • One real solution, where the two zeros are actually equal, i.e. the one $x$x-intercept where the graph of the quadratic function just touches the $x$x-axis at the turning point.
  • No real solutions, meaning the graph of the quadratic function has no $x$x-intercepts or real zeros.  

 

An aside

If we were to study this area of mathematics further, we would eventually learn of a theorem about polynomials that says for a polynomial of degree $n$n, there will be $n$n solutions. Since quadratics are polynomials of degree $2$2, we should always be expecting two solutions. So how is it that we can have one or even zero real solutions? 

The key here is the type of solutions we care about. At this stage we will be talking only about real solutions - solutions that are real numbers. But there are other types of numbers that are not real, in fact they are called imaginary numbers. (An unfortunately confusing name, they are just as real as real numbers!)

The set of real numbers and the set of imaginary numbers combine to make up the set of complex numbers, and it is the complex numbers that are ultimately the roots of polynomials. This is a very exciting area of maths, but we'll save that for later. For now let's get back to reality!

 

The discriminant - finding the number of solutions

Finding the number of solutions to a quadratic, or if there are any solutions at all, can be done without having to work through all the algebra required to solve the function. 

Let's look again at the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=b±b24ac2a

Specifically, let's look at what happens if the square root part $\sqrt{b^2-4ac}$b24ac takes on different values...

$b^2-4ac<0$b24ac<0  or  $b^2-4ac=0$b24ac=0  or  $b^2-4ac>0$b24ac>0 

If $b^2-4ac=0$b24ac=0, then the square root is $0$0 and then the quadratic equation becomes just $x=\frac{-b}{2a}$x=b2a. Does this look familiar? It is actually the equation for the axis of symmetry. It is just one number so there is just one real root (solution).

If $b^2-4ac>0$b24ac>0, then the square root will have two values,  $+$+$\sqrt{b^2-4ac}$b24ac and $-\sqrt{b^2-4ac}$b24ac. The quadratic formula will then generate for us two distinct real roots.

If $b^2-4ac<0$b24ac<0, then the square root is negative, and we know that we cannot take the square root of a negative number and get real solutions. This is the case where we have zero real roots.

This expression $b^2-4ac$b24ac within the quadratic formula is called the discriminant, and it determines the number of real solutions a quadratic function will have.

 

Discriminant of a quadratic

$b^2-4ac=0$b24ac=0, $1$1 real solution, $2$2 equal real roots, the quadratic just touches the $x$x-axis (it looks like it bounced off)

$b^2-4ac>0$b24ac>0, $2$2 real solutions, $2$2 distinct real roots, the quadratic passes through two different points on the $x$x-axis

$b^2-4ac<0$b24ac<0, $0$0 real solutions, $2$2 complex roots, the quadratic has no $x$x-intercepts

 

Rational or irrational solutions?

Remember that square roots of perfect squares like $\sqrt{9}$9 are rational because they can be written as fractions such as $\frac{3}{1}$31. Numbers like $\sqrt{7}$7 are irrational because they cannot be written as a fraction of whole numbers. Therefore if the discriminant is equal to a perfect square we know the quadratic equation has rational solutions. If it is not equal to a perfect square then the solutions must be irrational. 

 

Practice questions

Question 3

Consider the equation $2x^2-2x=x-1$2x22x=x1.

  1. Find the value of the discriminant.

  2. Using your answer from the previous part, determine the number of real solutions the equation has.

    No real solutions

    A

    Two real solutions

    B

    One real solution

    C

Question 4

Consider the equation $x^2+22x+121=0$x2+22x+121=0.

  1. Find the value of the discriminant.

  2. Using your answer from the previous part, determine whether the solutions to the equation are rational or irrational.

    Irrational

    A

    Rational

    B

Question 5

Consider the equation $x^2+18x+k+7=0$x2+18x+k+7=0.

  1. Find the values of $k$k for which the equation has no real solutions.

  2. If the equation has no real solutions, what is the smallest integer value that $k$k can have?

 

Outcomes

1.1.9

solve quadratic equations including the use of the quadratic formula and completing the square

1.1.11

determine turning points and zeros of quadratics and understand the role of the discriminant

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