Solving Equations with a Quadratic Formula: A Comprehensive Guide

Solving systems of equations is a fundamental skill in algebra, and there are various methods to find the values of variables that satisfy all given equations. In this guide, we will explore how to solve a system of two equations, ( x - y 10 ) and ( xy 75 ), step by step. We will focus on how to use the quadratic formula to solve for one variable in terms of the other, and then substitute back into the equations to find all possible solutions.

Understanding the Problem

We are given the following system of equations:

Equation 1: x - y 10

Equation 2: xy 75

Solving for One Variable in Terms of the Other

From Equation 1, we can express x in terms of y:

x y 10

Next, we substitute this expression for x into Equation 2:

(y 10)y 75

Expanding and simplifying, we get:

y^2 10y 75

y^2 10y - 75 0

Applying the Quadratic Formula

The quadratic formula is given by:

y frac{-b pm sqrt{b^2 - 4ac}}{2a}

Here, a 1, b 10, and c -75. Substituting these values, we get:

y frac{-10 pm sqrt{10^2 - 4 cdot 1 cdot (-75)}}{2 cdot 1}

y frac{-10 pm sqrt{100 300}}{2}

y frac{-10 pm sqrt{400}}{2}

y frac{-10 pm 20}{2}

This gives us two possible solutions for y:

y frac{10}{2} 5 quad text{or} quad y frac{-30}{2} -15

Substituting Back to Find the Values of x

Now, we substitute these values of y back into the expression x y 10 to find the corresponding values of x:

If y 5, then:

x 5 10 15

If y -15, then:

x -15 10 -5

Therefore, the possible values for x are 15 and -5.

Verification

To verify our solutions, we can substitute these values of x and y back into the original equations:

If x 15 and y 5, then:

Equation 1: 15 - 5 10 (true)

Equation 2: 15 * 5 75 (true)

If x -5 and y -15, then:

Equation 1: -5 - (-15) 10 (true)

Equation 2: (-5) * (-15) 75 (true)

So, the solutions are confirmed to be correct.

Alternative Method

Another way to solve these equations is to express xy 75 in a different form:

xy pmsqrt{(x - y)^2 times 4xy}

xy pmsqrt{(10^2 times 4 times 75)}

xy pmsqrt{400}

xy pm 20

Then, we solve for x using:

x frac{xy}{y} frac{pm 20}{y}

Substituting ( y 5 ) and ( y -15 ) into the equation for x, we get:

When y 5, x frac{20}{5} 15

When y -15, x frac{20}{-15} -5

This confirms our previous solutions.

Conclusion

In conclusion, the possible values for x in the given system of equations are 15 and -5. By using the quadratic formula and substituting back into the equations, we were able to find these values and verify them as correct. This method is a powerful tool for solving systems of equations involving quadratic expressions.

FAQs

What is the quadratic formula? The quadratic formula is a formula used to solve quadratic equations. It is expressed as y frac{-b pm sqrt{b^2 - 4ac}}{2a}, where a, b, and c are coefficients of a quadratic equation in the form ay^2 by c 0. How can I check if my solutions are correct? Substitute the values of x and y back into the original equations to ensure they satisfy both. If they do, then your solutions are correct. Are there other methods to solve this system of equations? Yes, there are other methods such as substitution, elimination, and graphing. The quadratic formula method provides a systematic approach, particularly useful when dealing with more complex quadratic expressions.