Problem 46 - Algebra 2

Question

Solve each equation. Check your answers. $$ -5 x^{2}-3=0 $$

Step-by-Step Solution

Answer

The equation has no real solutions, as the square root of a negative number is imaginary.

Step 1: Isolate the quadratic term

Add 3 to both sides of the equation to isolate the quadratic term: $$-5 x^{2} - 3 + 3 = 0 + 3$$$$-5 x^{2} = 3$$

Step 2: Divide both sides by -5

Divide both sides of the equation by -5 to solve for x squared: $$\frac{-5 x^{2}}{-5} = \frac{3}{-5}$$$$x^{2} = -\frac{3}{5}$$

Step 3: Solve for x by taking square roots

Take the square root of both sides of the equation to solve for x. Remember to include both the positive and negative roots: $$x = \pm\sqrt{-\frac{3}{5}}$$Since the square root of a negative number is not a real number, there are no real solutions.

Key Concepts

Quadratic Term Isolation

Quadratic term isolation is the first crucial step in solving quadratic equations. It involves rearranging the equation so that the term with the highest degree—that is, the term with the squared variable—is on one side of the equation by itself. In our example, the equation is \( -5x^2 - 3 = 0 \). To isolate the quadratic term \( x^2 \) on one side, we need to move the constant term \( -3 \) to the other side by adding \( 3 \) to both sides. This leaves us with \( -5x^2 = 3 \).

Isolating the quadratic term is a fundamental step because it sets up the equation for further manipulation and eventually reaching a solution. This process is similar to balancing a scale; whatever you do to one side, you must do to the other to maintain the equation's integrity.

Solving by Taking Square Roots

Once you have isolated the quadratic term, as we did with \( -5x^2 = 3 \), the next step to find the value of \( x \) is solving by taking square roots. To do this, we first need to get \( x^2 \) on its own. This is achieved by dividing both sides by \( -5 \), which gives us \( x^2 = -\frac{3}{5} \).

The principle of solving by taking square roots is to undo the squaring. Generally, you would take the square root of both sides. However, it is crucial to remember that every positive number has two square roots: one positive and one negative. Therefore, you denote the solution as \( x = \pm\sqrt{-\frac{3}{5}} \) to indicate both possible roots. However, this case is peculiar because you are asked to take the square root of a negative number, which leads us to the concept of no real solutions.

No Real Solutions

In our example, the equation led us to \( x = \pm\sqrt{-\frac{3}{5}} \). The square root of a negative number does not result in a real number. Real numbers are those that can be found on the number line, including both positive and negative numbers, as well as zero. They do not include imaginary or complex numbers, which is what the square root of a negative number is. Since we are dealing with the domain of real numbers in most basic algebra courses, the presence of a negative number inside a square root signifies that there are no real solutions to the equation. This is an essential concept as it shows the limitations within real number solutions and opens the doorway to exploring the field of complex numbers for further study in mathematics.

The understanding of these concepts is instrumental in mathematics as it equips students with the knowledge to discern when equations have real solutions, how to correctly isolate and manipulate terms, and the significance of each step while solving a quadratic equation.