How to Solve a Quadratic Equation: 3 Methods
What Is a Quadratic Equation?
A quadratic equation is any equation that can be rearranged into the form ax^2+bx+c=0 where a, b and c are numbers with a ≠ 0 (if a = 0, we get a linear equation bx+c=0).
To solve a quadratic, we are looking for the values of x, which make ax^2+bx+c=0 true. There will always be either two real solutions, two complex number solutions or exactly one solution (this is known as a double root).
In this article, we will look at the three most popular algebraic methods for finding these solutions.
To solve a quadratic, we are looking for the values of x, which make ax^2+bx+c=0 true. There will always be either two real solutions, two complex number solutions or exactly one solution (this is known as a double root).
In this article, we will look at the three most popular algebraic methods for finding these solutions.
Method 1: Factorising the Equation
If the quadratic equation has real, rational solutions, the quickest way to solve it is often to factorise into the form (px + q)(mx + n), where m, n, p and q are integers. This is especially true where the coefficient of x^2 is 1.
Example 1: Solve x^2+7x+12=0
To factorise, we are looking for two numbers that multiply to make 12 and add to make 7. This is 3 and 4, so we get x^2+7x+12=(x+3)(x+4) = 0.
We now have two brackets multiplying to make 0; hence our solutions must be when these brackets each equal 0. Our solutions are, therefore, x=−3 and x=−4.
Example 2: Solve 6x^2+11x−10=0
A slightly trickier one to factorise, but by first considering the 6 and −10 and with some trial and error, we can see that 6x^2+11x−10=(2x+5)(3x−2)=0. Again, we can see that the brackets must equal zero, so x=−5/2 and x=2/3.
Example 3: Solve x^2−36=0
An expansion of this method is when we have the difference of two squares. In this example, we have the square of x minus the square of 6. It can be seen quite quickly that x^2−36 = (x+6)(x−6), and therefore, we get the answers −6 and 6.
Example 1: Solve x^2+7x+12=0
To factorise, we are looking for two numbers that multiply to make 12 and add to make 7. This is 3 and 4, so we get x^2+7x+12=(x+3)(x+4) = 0.
We now have two brackets multiplying to make 0; hence our solutions must be when these brackets each equal 0. Our solutions are, therefore, x=−3 and x=−4.
Example 2: Solve 6x^2+11x−10=0
A slightly trickier one to factorise, but by first considering the 6 and −10 and with some trial and error, we can see that 6x^2+11x−10=(2x+5)(3x−2)=0. Again, we can see that the brackets must equal zero, so x=−5/2 and x=2/3.
Example 3: Solve x^2−36=0
An expansion of this method is when we have the difference of two squares. In this example, we have the square of x minus the square of 6. It can be seen quite quickly that x^2−36 = (x+6)(x−6), and therefore, we get the answers −6 and 6.
Method 2: Completing the Square
A similar method to above, completing the square also involves factorising, but this method will also work for complex and/or irrational answers. This time, we are looking to convert our quadratic into the form (x+q)^2+r=0 where q and r are real numbers to be found. Let's start with the same example as above.
Example 1: Solve x^2+7x+12=0
Consider (px+q)^2. When we expand this, we get p^2 x^2+2pqx+q^2. In our example, the coefficient of x^2 is 1, so p=1. The coefficient of x is 7, so 2pq=7 and hence q = 7/2.
Our expression must, therefore, begin (x+7/2)^2. If we expand this we get (x+7/2)^2 = x^2+7x+49/4 and so, by rearranging we can see that x^2+7x = (x+7/2)^2 − 49/4.
Therefore x^2+7x−12 = (x+7/2)^2 − 49/4 + 12
=(x+7/2)^2 − 1/4.
Setting this equal to 0 and rearranging, we get:
(x+7/2)^2 − 1/4 = 0
(x+7/2)^2 = 1/4
x+7/2 = ±1/2
x = −7/2 ±1/2
x = −3 and −4
Example 2: Solve 2x^2−3x−4=0
This time, we have a coefficient of x^2, which is not 1. In order to complete the square on these types of quadratics, it is easiest to first divide by this coefficient. In this case, dividing by 2 gives us:
x^2−(3/2)x−2=0
By our method from above, we put half of 3/2 into the bracket and then remove the square of this to cancel out the extra part given when the bracket is expanded, so:
x^2−(3/2)x−2=(x−3/4)^2−(3/4)^2−2
=(x−3/4)^2−41/16
Setting this equal to 0 and rearranging gives us:
(x−3/4)^2−41/16=0
(x−3/4)^2=41/16
x−3/4=±(√41)/4
x=3/4 ±(√41)/4 (approximately −0.851 and 2.351)
Example 1: Solve x^2+7x+12=0
Consider (px+q)^2. When we expand this, we get p^2 x^2+2pqx+q^2. In our example, the coefficient of x^2 is 1, so p=1. The coefficient of x is 7, so 2pq=7 and hence q = 7/2.
Our expression must, therefore, begin (x+7/2)^2. If we expand this we get (x+7/2)^2 = x^2+7x+49/4 and so, by rearranging we can see that x^2+7x = (x+7/2)^2 − 49/4.
Therefore x^2+7x−12 = (x+7/2)^2 − 49/4 + 12
=(x+7/2)^2 − 1/4.
Setting this equal to 0 and rearranging, we get:
(x+7/2)^2 − 1/4 = 0
(x+7/2)^2 = 1/4
x+7/2 = ±1/2
x = −7/2 ±1/2
x = −3 and −4
Example 2: Solve 2x^2−3x−4=0
This time, we have a coefficient of x^2, which is not 1. In order to complete the square on these types of quadratics, it is easiest to first divide by this coefficient. In this case, dividing by 2 gives us:
x^2−(3/2)x−2=0
By our method from above, we put half of 3/2 into the bracket and then remove the square of this to cancel out the extra part given when the bracket is expanded, so:
x^2−(3/2)x−2=(x−3/4)^2−(3/4)^2−2
=(x−3/4)^2−41/16
Setting this equal to 0 and rearranging gives us:
(x−3/4)^2−41/16=0
(x−3/4)^2=41/16
x−3/4=±(√41)/4
x=3/4 ±(√41)/4 (approximately −0.851 and 2.351)
Method 3: The Quadratic Formula
Our third method works for all quadratic equations, whether their solutions are rational or irrational, real or complex. The quadratic formula is a set formula which can be derived by completing the square on the quadratic ax^2+bx+c=0. To see how this is done, watch the video below.
For any quadratic equation of the form ax^2+bx+c=0, the solutions can be found by using the formula x = (−b ± √(b^2 − 4ac))/2a.
Example 1: Solve x^2+7x+12=0
Using the quadratic formula with a = 1, b = 7 and c = 12:
x = (−7 ± √(7^2 − 4 × 1 × 12)) / (2 × 1)
= (−7 ± √(49 − 48)) / 2
= (−7 ± √1) / 2
= (−7 ± 1) / 2
= −4 or −3
Example 2: Solve 2x^2−3x−4=0
x = (3 ± √((−3)^2 − 4 × 2 × −4)) / (2 × 2)
= (3 ± √(9 + 32)) / 4
= (3 ± √41) / 4
= 3/4 + (√41)/4
Example 1: Solve x^2+7x+12=0
Using the quadratic formula with a = 1, b = 7 and c = 12:
x = (−7 ± √(7^2 − 4 × 1 × 12)) / (2 × 1)
= (−7 ± √(49 − 48)) / 2
= (−7 ± √1) / 2
= (−7 ± 1) / 2
= −4 or −3
Example 2: Solve 2x^2−3x−4=0
x = (3 ± √((−3)^2 − 4 × 2 × −4)) / (2 × 2)
= (3 ± √(9 + 32)) / 4
= (3 ± √41) / 4
= 3/4 + (√41)/4
Comments
Which of these methods is your favourite?
Don't forget to leave your comments below.
Don't forget to leave your comments below.