Question

Solve the quadratic equation by extracting square roots. List both the exact answer and a decimal answer that has been rounded to two decimal places. $$ 5 x^{2}=190 $$

Short answer

The solutions to the equation are \( x = \sqrt{38} \), \( x = -\sqrt{38} \), or in decimal form \( x \approx \pm 6.16 \).

Step-by-step solution

Isolate the \( x^2 \) term.

Start by isolating the \( x^2 \) term by dividing both sides of the equation by \( 5 \): \[ x^2 = \frac{190}{5} = 38 \]

Extract Square

Next, take the square root of both sides to solve for \( x \). Remember that square root has two solutions, one positive and one negative: \[ x = \sqrt{38} \] and \[ x = -\sqrt{38} \]

Find Decimal Approximations

Finally, calculate the decimal approximation of the solutions to two decimal places: \[ x \approx \pm 6.16 \]

Key concepts

Extracting Square Roots

When solving quadratic equations, one intuitive method is extracting square roots. This approach is applicable when the equation can be simplified to the form \( x^2 = k \) where \( k \) is a non-negative real number. To extract square roots, you apply the square root operation to both sides of the equation. Since \( x^2 \) represents a squared term, taking the square root is essentially reverse-engineering this operation to find \( x \) itself.

However, it's crucial to consider that there are always two possible solutions for \( x \) because squaring either a positive or negative number results in a positive outcome. Hence, when you apply the square root to both sides, you include both the positive and negative roots. In mathematical terms, if \( x^2 = k \) then \( x = \(\pm\)\sqrt{k} \) where \( + \) denotes the positive solution and \( - \) signifies the negative solution. Remember that if \( k \) is negative, \( x \) would involve imaginary numbers, but for most basic quadratic equations, \( k \) will be positive.

Quadratic Formula

The quadratic formula provides a fail-safe way to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is as follows: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this powerful technique, \( a \) refers to the coefficient of \( x^2 \), \( b \) refers to the coefficient of \( x \), and \( c \) is the constant term. The expression under the square root, \( b^2 - 4ac \), is known as the discriminant and it determines the nature of the solutions. If the discriminant is positive, the equation has two distinct real number solutions; if it's zero, there is one real solution; and if it's negative, there are two complex solutions.

Using the quadratic formula can be especially helpful when other methods like factoring or completing the square are not feasible or when you have to deal with coefficients that are not perfect squares. It's a versatile technique that allows you to always find the exact solutions to a quadratic equation.

Decimal Approximations

Decimal approximations are used when we need to express solutions of an equation in a more practical, usable form. Often, the exact solutions to equations involving square roots, like \( \sqrt{38} \), are irrational numbers—numbers that cannot be expressed as a simple fraction. For most real-world applications, we round these numbers to a certain number of decimal places.

It's important to use a consistent method for rounding. For example, rounding to two decimal places involves looking at the third decimal digit. If this digit is 5 or more, we increase the second decimal digit by 1; if it's less than 5, we leave the second digit as is. In the case of \( \sqrt{38} \), the square root of 38 to two decimal places is approximately 6.16, denoting that the third decimal place was 5 or higher. Remember, when approximating decimal solutions, you should always provide a context for the approximation, such as the required precision in a real-world scenario, to indicate its limitations and accuracy.