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Solve rational equations that result in quadratics


Sometimes when we remove the denominators in fractional equations the resulting equation is a quadratic.

For example:

$\frac{6+7x}{3x}$6+7x3x $=$= $x$x
$6+7x$6+7x $=$= $x\times3x$x×3x
$6+7x$6+7x $=$= $3x^2$3x2
$6$6 $=$= $3x^2-7x$3x27x
$0$0 $=$= $3x^2-7x-6$3x27x6

Now it's more obvious that the equation we have to solve is a quadratic. From here we choose one of our solving options:

  • Solve by factorisation
  • Solve by completing the square
  • Solve by using the quadratic formula

Let's use the quadratic formula for this one.


Example 1

For the equation $y=3x^2-7x-6$y=3x27x6 we can see that $a=3$a=3$b=-7$b=7 and $c=-6$c=6.

$x$x $=$= $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$b±b24ac2a
$x$x $=$= $\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4\times3\times\left(-6\right)}}{2\times3}$(7)±(7)24×3×(6)2×3
$x$x $=$= $\frac{7\pm\sqrt{49+72}}{6}$7±49+726
$x$x $=$= $\frac{7\pm\sqrt{121}}{6}$7±1216
$x$x $=$= $\frac{7\pm11}{6}$7±116
$x$x $=$= $\frac{18}{6},-\frac{4}{6}$186,46
$x$x $=$= $3,-\frac{2}{3}$3,23


The real key to solving these types of quadratics is not in the solving of the actual quadratic but in getting it to the quadratic form in the first place. This means we need to understand how to deal with fractions!


Example 2

Here is another example, $\frac{15}{x}-3x=-4$15x3x=4.

To remove the denominator of $x$x in the first term, we must multiply ALL terms by the factor of $x$x.

Hence $\frac{15}{x}-3x=-4$15x3x=4, becomes $\frac{15}{x}\times x-3x\times x=-4\times x$15x×x3x×x=4×x

which is $15-3x^2=-4x$153x2=4x. From here move all terms onto one side and solve. 


Example 3

The fractions can get pretty messy. Take this example, $\frac{3}{x-2}+\frac{2}{x+2}=\frac{6}{x}$3x2+2x+2=6x.

Looks a bit nasty doesn't it. Just take it a step at a time.

Let's multiply all terms by the factor $(x-2)$(x2) first.

$\frac{3}{x-2}+\frac{2}{x+2}$3x2+2x+2 $=$= $\frac{6}{x}$6x
$\frac{3}{x-2}\times(x-2)+\frac{2}{x+2}\times(x-2)$3x2×(x2)+2x+2×(x2) $=$= $\frac{6}{x}\times(x-2)$6x×(x2)
$3+\frac{2(x-2)}{x+2}$3+2(x2)x+2 $=$= $\frac{6(x-2)}{x}$6(x2)x

Now, we multiply all terms by $(x+2)$(x+2)

$3+\frac{2(x-2)}{x+2}$3+2(x2)x+2 $=$= $\frac{6(x-2)}{x}$6(x2)x
$3(x+2)+2(x-2)$3(x+2)+2(x2) $=$= $\frac{6(x-2)(x+2)}{x}$6(x2)(x+2)x
$3x+2+2x-4$3x+2+2x4 $=$= $\frac{6(x^2-4)}{x}$6(x24)x
$5x-2$5x2 $=$= $\frac{6x^2-24}{x}$6x224x

Just one last fractional piece to take care of, the $x$x on the right. So multiply all terms by $x$x.

$5x-2$5x2 $=$= $\frac{6x^2-24}{x}$6x224x
$5x^2-2x$5x22x $=$= $6x^2-24$6x224
$0$0 $=$= $x^2+2x-24$x2+2x24

So the key is to take it one step at a time, deal with each piece individually and be careful with our multiplication, expanding and of course watch for negative signs!

Worked Examples


Solve the following equation:


  1. Write all solutions on the same line, separated by commas.


Solve for the unknown.


  1. Write all solutions on the same line, separated by commas.


Solve the equation $\frac{7}{x\left(x+3\right)}+5=\frac{x+4}{x}$7x(x+3)+5=x+4x.

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