{"id":1348,"date":"2024-05-10T02:36:50","date_gmt":"2024-05-10T02:36:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1348"},"modified":"2024-11-25T18:14:49","modified_gmt":"2024-11-25T18:14:49","slug":"other-types-of-equations-apply-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/other-types-of-equations-apply-it-2\/","title":{"raw":"Other Types of Equations: Apply It 2","rendered":"Other Types of Equations: Apply It 2"},"content":{"raw":"

Solving Rational Equations That Lead to Quadratics<\/strong><\/h2>\r\nWe've looked at how to solve rational equations before. Sometimes, when you're working through a rational equation, you end up with a quadratic equation. If that happens, you can simplify and solve the quadratic using the methods we've learned, like factoring or using the quadratic formula. Keep in mind that sometimes, after solving, you might find that there are no valid solutions.\r\n\r\n
Solve the following rational equation:\r\n

[latex]\\dfrac{-4x}{x - 1}+\\dfrac{4}{x+1}=\\dfrac{-8}{{x}^{2}-1}[\/latex]<\/p>\r\n\r\n\r\n

\r\n\r\nRecall an important concept in working with rational expressions: To add or subtract fractions, we first need to find a common denominator<\/strong>. This ensures that the fractions have the same base, allowing us to combine them effectively.\r\n\r\nLet's write the equation with this common denominator and simplify the terms:\r\n
\r\n
[latex]\\begin{align*} \\text{Original equation} & : & \\frac{-4x}{x-1} + \\frac{4}{x+1} &= \\frac{-8}{x^2 - 1} \\\\ \\text{Recognize common denominator} & : & x^2 - 1 &= (x-1)(x+1) \\\\ \\text{Rewrite with common denominator} & : & \\\\ \\text{Multiply the first term by } \\frac{x+1}{x+1} & : & \\frac{-4x(x+1)}{x^2-1} &= \\frac{-4x^2 - 4x}{x^2-1} \\\\ \\text{Multiply the second term by } \\frac{x-1}{x-1} & : & \\frac{4(x-1)}{x^2-1} &= \\frac{4x - 4}{x^2-1} \\\\ \\text{Combine terms over common denominator} & : & \\frac{-4x^2 - 4x + 4x - 4}{x^2-1} &= \\frac{-8}{x^2-1} \\\\ \\text{Simplify numerator} & : & \\frac{-4x^2 - 4}{x^2-1} &= \\frac{-8}{x^2-1} \\\\ \\end{align*}[\/latex]<\/span><\/div>\r\n<\/div>\r\nNow that we have matching denominators, we equate the numerators:\r\n\r\n
[latex]\\begin{align*} \\text{Equate numerators} & : & -4x^2 - 4 &= -8 \\\\ \\text{Simplify} & : & -4x^2 - 4 + 8 &= 0 \\\\ & : & -4x^2 + 4 &= 0 \\\\ \\text{Divide by -4} & : & x^2 - 1 &= 0 \\\\ \\text{Factor} & : & (x-1)(x+1) &= 0 \\\\ \\end{align*}[\/latex]<\/center>Finally, solve for [latex]x[\/latex]:\r\n\r\n
[latex]\\begin{align*} x-1 &= 0 & x+1 &= 0 \\\\ x &= 1 & x &= -1 \\end{align*}[\/latex]<\/center>Lastly, check for extraneous solution:\r\n\r\nRemember: we must check for extraneous solutions because the original equation had denominators that could be undefined for certain values of [latex]x[\/latex].\r\n\r\nBoth [latex]x=1[\/latex] and [latex]x=-1[\/latex] make the original denominators zero, so they are not valid solutions.\r\n\r\nTherefore, this equation actually has no solution<\/strong>.\r\n\r\n<\/section>
Solve the rational equation:\r\n

[latex]\\dfrac{3x+2}{x - 2}+\\dfrac{1}{x}=\\dfrac{-2}{{x}^{2}-2x}[\/latex]<\/p>\r\n[reveal-answer q=\"662482\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"662482\"]\r\n\r\n[latex]\\begin{align*} \\text{Original equation} & : \u00a0 \\frac{3x+2}{x - 2} + \\frac{1}{x} = \\frac{-2}{x^2-2x} \\\\ \\text{Factor the denominator on the right-hand side}\u00a0 & : \u00a0 x^2 - 2x = x(x-2) \\\\ \\text{Rewrite each fraction with the common denominator } x(x-2) & : \\\\ \\frac{3x+2}{x-2} = \\frac{(3x+2)x}{x(x-2)} \\\\ \\frac{1}{x} = \\frac{x-2}{x(x-2)} \\\\ \\text{Combine and simplify the numerators} & : \\frac{3x^2 + 2x + x - 2}{x(x-2)} = \\frac{-2}{x(x-2)} \\\\ \\text{Simplify further} & : \\frac{3x^2 + 3x - 2}{x(x-2)} = \\frac{-2}{x(x-2)} \\\\ \\text{Set the numerators equal to each other} & : 3x^2 + 3x - 2 = -2 \\\\ \\text{Simplify this to form a quadratic equation} & : 3x^2 + 3x = 0 \\\\ \\text{Factor out } x \\text{ from the quadratic} & : x(3x + 3)= 0 \\\\ \\text{Solve for } x & : x = 0 \\quad \\text{or} \\quad 3x + 3 = 0 \\\\ \\text{If } 3x + 3 = 0 & : x = -1 \\end{align*}[\/latex]\r\n\r\nCheck for extraneous solution:<\/strong>\r\n\r\nBased on the original denominators [latex]x(x-2)[\/latex], [latex]x \\ne 0[\/latex] and [latex]x \\ne 2[\/latex].\r\n\r\nThis means that [latex]x=0[\/latex] is an extraneous solution.\r\n\r\nThis leaves us with the solution: [latex]x=-1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

[ohm2_question hide_question_numbers=1]18983[\/ohm2_question]<\/section>
[ohm2_question hide_question_numbers=1]18984[\/ohm2_question]<\/section>","rendered":"

Solving Rational Equations That Lead to Quadratics<\/strong><\/h2>\n

We’ve looked at how to solve rational equations before. Sometimes, when you’re working through a rational equation, you end up with a quadratic equation. If that happens, you can simplify and solve the quadratic using the methods we’ve learned, like factoring or using the quadratic formula. Keep in mind that sometimes, after solving, you might find that there are no valid solutions.<\/p>\n

Solve the following rational equation:<\/p>\n

[latex]\\dfrac{-4x}{x - 1}+\\dfrac{4}{x+1}=\\dfrac{-8}{{x}^{2}-1}[\/latex]<\/p>\n

\n

Recall an important concept in working with rational expressions: To add or subtract fractions, we first need to find a common denominator<\/strong>. This ensures that the fractions have the same base, allowing us to combine them effectively.<\/p>\n

Let’s write the equation with this common denominator and simplify the terms:<\/p>\n

\n
[latex]\\begin{align*} \\text{Original equation} & : & \\frac{-4x}{x-1} + \\frac{4}{x+1} &= \\frac{-8}{x^2 - 1} \\\\ \\text{Recognize common denominator} & : & x^2 - 1 &= (x-1)(x+1) \\\\ \\text{Rewrite with common denominator} & : & \\\\ \\text{Multiply the first term by } \\frac{x+1}{x+1} & : & \\frac{-4x(x+1)}{x^2-1} &= \\frac{-4x^2 - 4x}{x^2-1} \\\\ \\text{Multiply the second term by } \\frac{x-1}{x-1} & : & \\frac{4(x-1)}{x^2-1} &= \\frac{4x - 4}{x^2-1} \\\\ \\text{Combine terms over common denominator} & : & \\frac{-4x^2 - 4x + 4x - 4}{x^2-1} &= \\frac{-8}{x^2-1} \\\\ \\text{Simplify numerator} & : & \\frac{-4x^2 - 4}{x^2-1} &= \\frac{-8}{x^2-1} \\\\ \\end{align*}[\/latex]<\/span><\/div>\n<\/div>\n

Now that we have matching denominators, we equate the numerators:<\/p>\n

[latex]\\begin{align*} \\text{Equate numerators} & : & -4x^2 - 4 &= -8 \\\\ \\text{Simplify} & : & -4x^2 - 4 + 8 &= 0 \\\\ & : & -4x^2 + 4 &= 0 \\\\ \\text{Divide by -4} & : & x^2 - 1 &= 0 \\\\ \\text{Factor} & : & (x-1)(x+1) &= 0 \\\\ \\end{align*}[\/latex]<\/div>\n

Finally, solve for [latex]x[\/latex]:<\/p>\n

[latex]\\begin{align*} x-1 &= 0 & x+1 &= 0 \\\\ x &= 1 & x &= -1 \\end{align*}[\/latex]<\/div>\n

Lastly, check for extraneous solution:<\/p>\n

Remember: we must check for extraneous solutions because the original equation had denominators that could be undefined for certain values of [latex]x[\/latex].<\/p>\n

Both [latex]x=1[\/latex] and [latex]x=-1[\/latex] make the original denominators zero, so they are not valid solutions.<\/p>\n

Therefore, this equation actually has no solution<\/strong>.<\/p>\n<\/section>\n

Solve the rational equation:<\/p>\n

[latex]\\dfrac{3x+2}{x - 2}+\\dfrac{1}{x}=\\dfrac{-2}{{x}^{2}-2x}[\/latex]<\/p>\n