Question

Solve the equation algebraically. Check the solutions graphically. $$\frac{1}{5} x^{2}=5$$

Short answer

The algebraic solution to the equation \(\frac{1}{5} x^{2}=5\) is \(x = 5\) and \(x = -5\). The graphical solution confirms these results.

Step-by-step solution

Solve for \(x^2\)

Firstly, you multiply both sides of the equation by 5 to isolate \(x^2\). This gives \(x^{2} = 25\).

Solve for \(x\)

In the second step, you take the square root of both sides of the equation to solve for \(x\). Don't forget that taking the square root of a number gives a positive and a negative solution. Thus, \(x = \sqrt{25}\) gives \(x = 5\) and \(x = -5\).

Check the solution graphically

Now, you can confirm these solutions visually by plotting the original equation and checking that the graph touches the x-axis at both \(x = 5\) and \(x = -5\). The original equation \(\frac{1}{5} x^{2}=5\) when isolated for \(y=\frac{1}{5} x^{2}\) produces a U-shaped curve, called a parabola, which intersects the horizontal line \(y=5\) at two points, corresponding to \(x = 5\) and \(x = -5\).

Key concepts

Algebraic Methods

Understanding algebraic methods is essential when you embark on solving quadratic equations. These types of equations typically have the general form of \( Ax^2 + Bx + C = 0 \) where \( A \) , \( B \) , and \( C \) are constants, and \( A \) is not zero.

To solve them algebraically, you can use several techniques, such as factoring, completing the square, or employing the quadratic formula. However, when the equation is already in a simplified form, like the textbook exercise \( \frac{1}{5} x^{2}=5 \), you can solve it by isolating \( x \). This can involve straightforward algebraic manipulation – in this case, multiplying both sides by 5 to eliminate the fraction and then extracting the square roots to find the actual values of \( x \).

Square Roots

Square roots are a fundamental concept when dealing with quadratic equations. When you're confronted with \( x^2 = n \), where \( n \) is a positive real number, the square root operation allows you to find \( x \) because it's the number that, when multiplied by itself, gives \( n \).

Keep in mind that there are always two square roots: the positive \( \sqrt{n} \) and the negative \( -\sqrt{n} \). This property is particularly important in our example, \( x^2 = 25 \), where taking the square root gives us two solutions, \( x = 5 \) and \( x = -5 \). This duality arises from the fact that both \( (5)^2 \) and \( (-5)^2 \) result in 25, reflecting the inherent symmetry in square root functions.

Graphical Solutions

Graphical solutions offer a valuable visual representation to confirm the solutions of quadratic equations. When you plot the equation on a coordinate grid, it appears as a curve named a parabola. The points where the curve intersects the x-axis represent the solutions to the equation.

For the exercise \( \frac{1}{5} x^{2}=5 \), you'd first rewrite it as \( y=\frac{1}{5} x^{2} \) and then graph it. The intercepts with \( y=5 \) give us the graphical evidence for our algebraic solutions, where we'll see the curve touching the x-axis at \( x = 5 \) and \( x = -5 \). It's always reassuring to have that graphical check because it confirms the algebra we've done is correct.

Parabolas

Parabolas are symmetrical, U-shaped graphs that are very common in quadratic equations. Their distinct shape is determined by the coefficient of the \( x^2 \) term of the equation. If you look at a parabola's orientation, the opening faces upwards if the coefficient is positive, as it is in our exercise \( \frac{1}{5} x^{2}=5 \).

A parabola's vertex is the highest or lowest point and lies on the axis of symmetry, which in a standard form equation \( y = Ax^2 + Bx + C \) is given by \( x = -\frac{B}{2A} \). The intersection points of the parabola with the x-axis are the real roots of the quadratic equation. In the case of our example, the parabola intersects the x-axis at \( x = 5 \) and \( x = -5 \) , matching our algebraic solution.