Quadratic equations are defined as equations of second degree where at least one variable or term is raised to a power of 2. Solving quadratic equations is done by determining the roots of the equation, also known as x-intercepts. These are the values of x at which the graph cuts through the x-axis, as seen below.
Parabola showing where the x-intercepts or roots are located
How do you solve quadratic equations?
Quadratic Equations are solved using one of the following methods:
Taking square roots
Taking square roots is a method that can be used to solve quadratic equations when there is only one \(x^2\) term in the equation. It is done by isolating the \(x^2\) term and then using a square root to solve the equation by finding the values of \(x\).
\( 3x^2 = 48 \).
Step 1: Isolate the squared variable.
\( 3x^2 = 48 \)
\(x^2 = 16\)
Step 2: Solve your quadratic equation by calculating the square root of both sides of the equation. Remember that you will have two solutions because the square root of a number can either be positive or negative.
\( \sqrt{x^2} = ±\sqrt{16} x = ±4 \)
Factoring
Factoring is when we determine the terms that need to be multiplied together to get a mathematical expression. Factoring quadratic equations can be done in the following ways:
Taking the greatest common factor
Taking the greatest common factor (GCF) is a factoring method where we determine the highest term that evenly divides into all other terms. Let's see how this method works with an example:
\( 14x^2 - 35x = 0\)
Step 1: Find the greatest common factor by identifying the numbers and variables that each term has in common.
\( 14x^2 = 2 \cdot 7\cdot x \cdot x \)
\( 35x = 5 \cdot 7 \cdot x\)
As we can see the factors that are common to both terms are 7 and x, therefore the GCF = 7x.
Step 2: Write out each term as a product of the greatest common factor and another factor, i.e. the two parts of the term. The other factor can be determined by dividing your term by your GCF.
\(\frac{14x^2}{7x} = 2x \therefore 14x^2 = (2x) \cdot 7x\)
\(\frac{35x}{7x} = 5 \therefore 35x = 7x \cdot 5\)
The ∴ symbol means "therefore".
Step 3: Having rewritten each term, rewrite the quadratic equation in the following form: ab+ac=0
\( 14x^2 - 35x = 7x \cdot (2x) - 7x(5)\)
Step 4: Apply the law of distributive property and factor out the greatest common factor.
\( 7x(2x) -7x(5) = 7x(2x-5)\)
Step 5: Equate the factored expression to 0 and find the x-intercepts.
\( x_1: \begin{split} 7x = 0 \\ x = 0 \end {split} \)
\( x_2: \begin{split}2x - 5= 0 \\ x = \frac{5}{2}\end {split}\)
Perfect square
The perfect square method is when we transform a perfect square trinomial,
\( a^2 + 2ab + b^2\) or \( a^2 - 2ab + b^2\) into a perfect square binomial, \( (a + b)^2 \) or \( (a - b)^2\) . Let's have a look at solving quadratic equations using this method:
\( 9x^2 - 12x +4\)
Step 1: Transform your equation from standard form, \( ax^2 + bx + c = 0\), into a perfect square trinomial, \( a^2 + 2ab + b^2\).
\( 9x^2 - 12x +4 = (3x)^2 -2(3x)(2) - 2^2\)
Step 2: Transform the perfect square trinomial into a perfect square binomial,\( (a + b)^2 \)
\((3x)^2 -2(3x)(2) - 2^2 = (3x-2)^2\)
Step 3: Calculate the value of the x-intercept by equating the perfect square binomial to 0 and solving for x.\( (3x - 2)^2 = 0\)\( \sqrt{(3x-2^2} = ±\sqrt{0} \)
\( 3x - 2 = 0\)
\( x = \frac {2}{3} \)
Grouping
Grouping is when we group terms that have common factors before factoring. Let's look at an example:
\( 2x^2 - 7x - 15\)
Step 1: List out the values of a, b and c.
\( a = 2, b = -7, c = -15\)
Step 2: Find the factors that when multiplied equal \(a \cdot c\) , and when added equal b. T where numbers that product ac and also add to b.
\( ac = -30, b = -7\)
\(1 \cdot 30 = 30\)
\(2 \cdot 15 = 30\)
\(3 \cdot 10 = 30 \qquad 3-10 = -7\)
\(5 \cdot 6 = 30\)
The two numbers are therefore 3 and -10, as they add to -7. The other factors of 30 cannot be arranged in any way that would make them equal to -7.
Step 3: Use these factors to rewrite the x-term (bx) in the original expression/equation.
\( 2x^2 - 7x - 15 = 2x^2 + 3x - 10x - 15\)
Step 4: Use grouping to factor the expression. Group the first two terms and the last two terms together, then pull out common factors from both groups and combine like terms.
\( \begin{split} (2x^2 + 3x) - (10x - 15) = x (2x + 3) - 5(2x + 3) \\ = (x-5)(2x-3)\end{split}\)
Step 5: Equate the factored expression to 0 and solve for the x-intercepts.
\( (x - 5) (2x - 3)\)
\( x_1: \begin{split} x - 5 = 0 \\ x = 5 \end {split} \)
\( x_2: \begin{split}2x + 3= 0 \\ x = \frac{-3}{2}\end {split}\)
Completing the square
Completing the square is when we change the standard form of the quadratic equation into a perfect square with an additional constant. This means changing \( ax^2 + bx + c = 0 \) into \(a(x + m)^2 + n \), where m is a real number and n is a constant. They are calculated in the following way: \( m = \frac{b}{2a}\) and \(n = c - \frac{b^2}{4a} \).
\( 3x^2 - 5x - 7 = 0 \)Step 1: List out the values of a, b and c.
\(a = 3, b = -5, c = -7 \)
Step 2: Calculate the value of m by using the following equation: \( m = \frac{b}{2a}\)
\( \begin{split}m = \frac{-5}{2(3)} \\ = -\frac{5}{6}\end{split} \)
Step 3: Calculate the value of n by using the following equation: \( n = c - \frac{b^2}{4a}\)
\(n = -7 - \frac {(-5)^2}{4(3)}\)
\(\begin{split} n = -7 - \frac {25}{12} \\ = - \frac{109}{12} \end{split}\)
Step 4: Substitute your calculated values and value of a into the following equation: \(a(x + m)^2 + n\)\(3(x - \frac {5}{6})^2 - \frac{109}{12} \)
Step 5: Equate your equation to 0 and thereby solve the equation.\(3(x - \frac{5}{6})^2 - \frac{109}{12} = 0\)
\(3(x - \frac{5}{6})^2 = \frac{109}{12}\)
\(x - \frac{5}{6}^2 = \frac{109}{36}\)
\(\sqrt{(x - \frac{5}{6}^2} = ±\sqrt{\frac{109}{36}}\)
\(x - \frac {5}{6} = \pm \frac {\sqrt{109}}{6}\)
\(x_{1,2} = \pm \frac {\sqrt{109}}{6} + \frac{5}{6}\)
\(x_{1} : x = \frac{5 + \sqrt{109}}{6} = 2.57\)
\(x_2: x = \frac {5-\sqrt{109}}{6} = -0.91 \)
Quadratic formula
The quadratic formula is a formula that uses the coefficients and constants in a quadratic equation to solve the equation by determining its x-intercepts/roots. The quadratic formula is used to solve quadratic equations that are very difficult to factor. The quadratic formula is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
\(b^2 - 4ac\)
is what we refer to as the discriminant. Depending on its sign, we know how many solutions the given quadratic equation has. It is represented by the following symbol:
- A positive discriminant means that the quadratic equation has two different real number solutions.
- A negative discriminant means that none of the solutions are real numbers.
Real numbers are numbers that can be identified on a timeline. For example, infinity is not a real number because it doesn't have a measurable size and therefore can't be identified on a number line. - A discriminant that is equal to 0 means that the quadratic equation has a repeated real number solution.
Graphical representation of what the discriminant shows
The following steps will show us how to solve a quadratic equation by using the quadratic formula:
\( x^2 - 7x + 12 = 0 \)
Step 1: List out the values of a, b and c.
\(a = 1, b = -7, c = 12 \)
Step 2: Calculate the value of the discriminant.
\(\begin{split} \Delta =(-7)^2 -4(1)(12) \\ = 1 \end{split}\)
Step 3: Substitute the values of a,b and c into the Quadratic Formula and solve for both roots/solutions.
\(\begin{align}x_{1} : x&= \frac{-b + \sqrt{\Delta}}{2a} \\ &=\frac{-(-7)+\sqrt{1}}{2(1)} \\ &= 4 \end{align}\)
\(\begin{align}x_{2} : x&= \frac{-b - \sqrt{\Delta}}{2a} \\ &=\frac{-(-7)-\sqrt{1}}{2(1)} \\ &= 3 \end{align}\)
Solving Quadratic Equations - Key takeaways
- Quadratic equations are solved by determining the roots of the equation.
- Taking square roots is a method that can be used to solve quadratic equations when there is only one \(x^2\) in the equation.
- Factoring is when we determine the terms that need to be multiplied together to get a mathematical expression and can be done by taking the greatest common factor, perfect square and grouping.
- Completing the square is when we change the standard form of the quadratic equation into a perfect square with an additional constant.
- The quadratic formula is a method used to solve quadratic equations; \(x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
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