Explainer: Solving Quadratic Equations: Taking Square Roots

In this explainer, we will learn how to solve quadratic equations with no linear term using the square root property.

Definition: Quadratic Equation

A quadratic equation is an equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Žβ‰ 0.

When we solve a quadratic equation, we find the values of π‘₯ for which π‘Žπ‘₯+𝑏π‘₯+𝑐=0. A quadratic equation will have up to two different solutions for π‘₯. In this explainer, we will look at how to solve a simple quadratic which has no linear term, that is, a quadratic equation in the form π‘Žπ‘₯+𝑐=0.

Let us recall the definition of real numbers and rational and irrational numbers.

Definition: Real Numbers

A real number is a positive or negative number, including those which have decimal places. A real number can be rational or irrational. We can write real numbers as belonging to the set ℝ.

Definition: Rational and Irrational Numbers

A rational number is a number which can be written in the form π‘Žπ‘, where π‘Ž and 𝑏 are integers and 𝑏≠0. We can write rational numbers as belonging to the set β„š.

An irrational number is one which cannot be written in the form π‘Žπ‘, where π‘Ž and 𝑏 are integers.

To begin, let us consider the equation π‘₯=25.

If we wished to solve this for x, we would take the square root of both sides of the equation. We could solve as π‘₯=√25, so π‘₯=5. However, since βˆ’5Γ—βˆ’5 would also give an answer of 25, we can indicate the positive and negative roots here as π‘₯=±√25.

So, π‘₯=Β±5.

It is important to remember that the use of Β± indicates two solutions, so π‘₯=Β±5 represents both π‘₯=5 and π‘₯=βˆ’5.

Let us now look at some examples of solving quadratic equations by taking square roots.

Example 1: Solving a Quadratic Equation by Taking Square Roots

Solve (π‘₯+6)=4.

Answer

Starting from (π‘₯+6)=4, we take square roots of both sides of the equation. Recall that when taking square roots we need to consider both the positive and the negative roots. Therefore, π‘₯+6=±√4=Β±2.

Subtracting 6 from both sides gives us π‘₯=Β±2βˆ’6.

Therefore, π‘₯=+2βˆ’6π‘₯=βˆ’2βˆ’6.or

Hence, our answer is π‘₯=βˆ’4 or π‘₯=βˆ’8.

Example 2: Solving a Quadratic Equation by Taking Square Roots

Find the solution set of the equation 4π‘₯=π‘₯9.

Answer

To begin solving this, we first multiply both sides of the equation 4π‘₯=π‘₯9 by π‘₯, which gives 4=π‘₯9.

Next, we multiply both sides by 9, giving us 36=π‘₯.

Taking the square root of both sides, considering both the positive and the negative roots, we have ±√36=π‘₯Β±6=π‘₯.

Therefore, π‘₯=Β±6 and our answer is the set {6,βˆ’6}.

Example 3: Solving a Quadratic Equation by Taking Square Roots

Find the solution set of the equation (π‘₯βˆ’5)=100 in β„š.

Answer

We perform the same operations to both sides of the equation to solve for π‘₯. Taking the square root of both sides, considering both the positive and the negative roots, gives us (π‘₯βˆ’5)=100π‘₯βˆ’5=±√100π‘₯βˆ’5=Β±10.

We can now add 5 to both sides to isolate π‘₯, giving us π‘₯=5Β±10π‘₯=5βˆ’10π‘₯=5+10.or

Therefore, the solution set is {βˆ’5,15}.

We will now look at an example where we first need to rearrange the equation to collect terms together, before solving using square roots.

Example 4: Solving a Quadratic Equation by Taking Square Roots

Determine the solution set of the equation βˆ’2π‘₯+15=π‘₯βˆ’12.

Answer

To begin solving this equation, we collect all the like terms. Subtracting π‘₯ from both sides and then subtracting 15 from both sides give us βˆ’2π‘₯+15=π‘₯βˆ’12βˆ’3π‘₯+12=βˆ’12βˆ’3π‘₯=βˆ’27.

We can now divide both sides by βˆ’3, which gives us π‘₯=9. Taking the square root of both sides, considering both the positive and the negative roots, gives us π‘₯=±√9π‘₯=Β±3. Therefore, the solution set is {3,βˆ’3}.

Let us now look at an example where we can use the square root method to attempt to solve a quadratic but which gives no solutions for π‘₯.

Example 5: Solving a Quadratic Equation by Taking Square Roots

Determine the solution set of the equation 44π‘₯+9=0, given that π‘₯βˆˆβ„.

Answer

We can solve this by performing the same operations to both sides of the equation to solve for π‘₯.

Subtracting 9 from both sides of the equation and then dividing by 44 give us 44π‘₯+9=044π‘₯=βˆ’9π‘₯=βˆ’944.

We can now take the square root of both sides, considering both the positive and the negative roots, which gives us π‘₯=Β±ο„žβˆ’944.

However, neither π‘₯=ο„žβˆ’944 nor βˆ’ο„žβˆ’944 gives us a real solution, since each takes the square root of a negative value. Therefore, the solution set of the equation is the null set: πœ™.

In this example, we correctly followed the square root method, but there are no real solutions for π‘₯. If we consider the graph of the function 𝑓(π‘₯)=44π‘₯+9, having no solutions for π‘₯ simply means that the graph would not pass through the π‘₯-axis.

Key Points

  • We can use the square root method to solve a quadratic equation when the quadratic has no terms in π‘₯.
  • We can use the square root method by following the steps below.
    1. Collect the π‘₯-terms together, and collect the constant terms on the other side of the equation.
    2. Take the square root of both sides of the equation.
  • It is important to remember that when taking a square rootwe must consider both the positive and the negative values of the roots. Using the Β± sign is a useful way to indicate these two values.

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