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Magazine Chapter 1: Problem 93
Use the method of your choice to find all real solutions to each equation. See the strategy for solving ax\(^{2}+b x+c=0\) on page 71. $$\frac{x-12}{3-x}=\frac{x+4}{x+7}$$
Short Answer
Expert verified
The real solutions are x = 8 and x = -6.
Step by step solution
01
Set the equation
Given the equation: \[ \frac{x-12}{3-x} = \frac{x+4}{x+7} \] First, note it is a proportion of two fractions.
02
Cross-multiply
To solve the rational equation, use cross-multiplication: \[ (x-12)(x+7) = (x+4)(3-x) \]
03
Expand the expressions
Expand both products: \[ x^2 + 7x - 12x - 84 = 3x + 12 - x^2 - 4x \] Simplify each side: \[ x^2 - 5x - 84 = -x^2 - x + 12 \]
04
Combine like terms
Combine all terms on one side to get a single quadratic equation: \[ x^2 - 5x - 84 + x^2 + x - 12 = 0 \] Simplify: \[ 2x^2 - 4x - 96 = 0 \]
05
Factor the quadratic
Factor the equation: \[ 2(x^2 - 2x - 48) = 0 \] Simplify: \[ x^2 - 2x - 48 = 0 \]
06
Solve the quadratic
Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -2 \), and \( c = -48 \). Substitute into the quadratic formula: \[ x = \frac{2 \pm \sqrt{4 + 192}}{2} \] Simplifying: \[ x = \frac{2 \pm \sqrt{196}}{2} \] \[ x = \frac{2 \pm 14}{2} \] It yields: \[ x = 8 \] and \[ x = -6 \]
07
Check for extraneous solutions
Check both solutions in the original equation to verify they do not produce a zero denominator. For \( x = 8 \): \[ \frac{8-12}{3-8} = \frac{8-12}{-5} = \frac{-4}{-5} = \frac{4}{5} \] and \[ \frac{8+4}{8+7} = \frac{12}{15} = \frac{4}{5} \] This satisfies the equation.For \( x = -6 \): \[ \frac{-6-12}{3+6} = \frac{-18}{9} = -2 \] and \[ \frac{-6+4}{-6+7} = \frac{-2}{1} = -2 \] This also satisfies the equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross-multiplication in equations
When you're dealing with equations that have fractions, like \(\frac{x-12}{3-x} = \frac{x+4}{x+7}\), cross-multiplication is a super useful technique. It helps eliminate the fractions and makes solving the equation much easier.
Here鈥檚 how you can do it:
So, in our example: \((x-12)(x+7) = (x+4)(3-x)\). This step simplifies the equation and allows you to work with polynomials instead of fractions, setting you up for the next steps.
Here鈥檚 how you can do it:
- Take the numerator of one fraction and multiply it by the denominator of the other fraction.
- Do the same for the other numerator and denominator pair.
So, in our example: \((x-12)(x+7) = (x+4)(3-x)\). This step simplifies the equation and allows you to work with polynomials instead of fractions, setting you up for the next steps.
Factoring quadratics
Factoring quadratics is a way to simplify the quadratic equation to its root form so you can easily find its solutions. In our equation, we factored as follows:
First, simplify the expanded equation:
Factoring this, we need to find two numbers that multiply to -48 and add up to -2.
First, simplify the expanded equation:
- Combine terms to form \[ x^2 - 5x - 84 + x^2 + x - 12 = 0 \]
- Simplify to get \[ 2x^2 - 4x - 96 = 0 \] and then \[ x^2 - 2x - 48 = 0 \]
Factoring this, we need to find two numbers that multiply to -48 and add up to -2.
- The numbers that work are -8 and 6.
- So, we can write: \[ (x - 8)(x + 6) = 0 \] This step breaks the quadratic into two linear factors.
Quadratic formula
If factoring doesn't work or isn't straightforward, the quadratic formula is a powerful tool. This formula can solve any quadratic equation of the form \[ ax^2 + bx + c = 0 \].
The formula itself is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let鈥檚 break it down for our equation where \(a = 1\), \(b = -2\), and \(c = -48\):
This gives the solutions: \( x = 8 \) and \( x = -6 \).
The formula itself is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let鈥檚 break it down for our equation where \(a = 1\), \(b = -2\), and \(c = -48\):
- Calculate the discriminant: \[ b^2 - 4ac = (-2)^2 - 4(1)(-48) = 4 + 192 = 196 \]
- Take the square root of the discriminant: \[ \sqrt{196} = 14 \]
- Apply the values back into the quadratic formula: \[ x = \frac{2 \pm 14}{2} \]
This gives the solutions: \( x = 8 \) and \( x = -6 \).
Checking for extraneous solutions
After solving, it's crucial to check for extraneous solutions to ensure the solutions are valid in the original equation. Extraneous solutions can arise from steps like cross-multiplication.
In our example, we obtained two solutions, \( x = 8 \) and \( x = -6 \):
In our example, we obtained two solutions, \( x = 8 \) and \( x = -6 \):
- For \( x = 8 \):
Check if the values satisfy the original equation and do not make the denominators zero:- \(\frac{8-12}{3-8} = \frac{4}{5}\)
- \(\frac{8+4}{8+7} = \frac{4}{5}\)
- For \( x = -6 \):
- \(\frac{-6-12}{3+6} = -2\)
- \(\frac{-6+4}{-6+7} = -2\)
