# Solve the following equation: (x^2-2)/3+((x^2-1)/5)^2=7/9(x^2-2)?

May 3, 2016

$x = - \sqrt{11} , - \frac{\sqrt{19}}{3} , \frac{\sqrt{19}}{3} , \sqrt{11}$

This explanation gives a rather in-depth method of determining the steps to finding possible factors into which to rewrite a quadratic-type equation so that it is solvable without the quadratic equation and/or a calculator.

#### Explanation:

First square the term on the left hand side of the equation.

$\frac{{x}^{2} - 2}{3} + {\left({x}^{2} - 1\right)}^{2} / 25 = \frac{7}{9} \left({x}^{2} - 2\right)$

Expand the squared binomial. Recall that ${\left({x}^{2} - 1\right)}^{2} = \left({x}^{2} - 1\right) \left({x}^{2} - 1\right)$.

$\frac{{x}^{2} - 2}{3} + \frac{{x}^{4} - 2 {x}^{2} + 1}{25} = \frac{7}{9} \left({x}^{2} - 2\right)$

We can clear out the fractions by multiplying the equation by the least common denominator of $3 , 25 ,$ and $9 ,$ which is $225$.

Note that $225 = {3}^{2} \cdot {5}^{2}$, so $\frac{225}{3} = 75$, $\frac{225}{25} = 9$, and $\frac{225}{9} = 25$.

Multiplying through by $225$ gives:

$75 \left({x}^{2} - 2\right) + 9 \left({x}^{4} - 2 {x}^{2} + 1\right) = 25 \left(7\right) \left({x}^{2} - 2\right)$

Distribute each multiplicative constant.

$75 {x}^{2} - 150 + 9 {x}^{4} - 18 {x}^{2} + 9 = 175 {x}^{2} - 350$

Move all terms to one side and reorder the equation.

$9 {x}^{4} - 118 {x}^{2} + 209 = 0$

This has the potential to be factorable: the lack of ${x}^{3}$ and $x$ terms means that this may be able to be factored in the form $\left({x}^{2} + a\right) \left({x}^{2} + b\right)$.

To test for factors, note that we should find a pair of integers whose product is the product of the first and final coefficients, which is $9 \times 209 = {3}^{2} \cdot 11 \cdot 19$. The same integers whose product is ${3}^{2} \cdot 11 \cdot 19$ should have a sum of $- 118$.

Since the product is positive and sum is negative, we know both the integers will be positive.

The trick now is to find some combination of numbers that comes from ${3}^{2} \cdot 11 \cdot 19$ whose sum is $118$. (If we find the positive version, we can switch both numbers to their negative form easily.)

We should attempt to come up with groupings of the factors from ${3}^{2} \cdot 11 \cdot 19$ that don't exceed $118$.

We can preemptively eliminate the possibility of ${3}^{2} \cdot 19$ and $11 \cdot 19$ occurring as either one of our two integers, since both of these are greater than $118$. Thus, if we focus on $19$ since it is the largest factor, we know it will only exist as either $19$ or $3 \cdot 19$.

So, our only two options for the integers are:

$\left.\begin{matrix}\boldsymbol{\text{Integer 1" & " " & bb"Integer 2" & " " & bb"Sum" \\ 19 & " " & 3^2*11=99 & " " & 118 \\ 19*3=57 & " " & 3*11=33 & " }} & 90\end{matrix}\right.$

Hence our pair of numbers whose product is ${3}^{2} \cdot 11 \cdot 19$ and sum is $118$ is $19$ and $99$.

From this we can write the quartic as:

$9 {x}^{4} - 118 {x}^{2} + 209 = 9 {x}^{4} - 99 {x}^{2} - 19 {x}^{2} + 209$

Factor by grouping:

$9 {x}^{2} \left({x}^{2} - 11\right) - 19 \left({x}^{2} - 11\right) = \left(9 {x}^{2} - 19\right) \left({x}^{2} - 11\right) = 0$

Split this into two equations:

$9 {x}^{2} - 19 = 0 \text{ "=>" "x^2=19/9" "=>" } x = \pm \frac{\sqrt{19}}{3}$

${x}^{2} - 11 = 0 \text{ "=>" "x^2=11" "=>" } x = \pm \sqrt{11}$

May 3, 2016

Equations with fractions always look worse than they are. As long as you have an equation and not an expression, you can get rid of the denominators by multiplying through by the LCM of denominators.

#### Explanation:

$\frac{{x}^{2} - 2}{3} + {\left(\frac{{x}^{2} - 1}{5}\right)}^{2} = \frac{7}{9} \left({x}^{2} - 2\right)$

Let's start by squaring the denominator in the second term.

$\frac{{x}^{2} - 2}{3} + \frac{{\left({x}^{2} - 1\right)}^{2}}{25} = \frac{7}{9} \left({x}^{2} - 2\right)$

Now multiply each term by 225 to cancel the denominators.
${\cancel{225}}^{75} \times \frac{\left({x}^{2} - 2\right)}{\cancel{3}} + {\cancel{225}}^{9} \frac{{\left({x}^{2} - 1\right)}^{2}}{\cancel{25}} = {\cancel{225}}^{25} \times \frac{7}{\cancel{9}} \left({x}^{2} - 2\right)$

$75 \left({x}^{2} - 2\right) + 9 {\left({x}^{2} - 1\right)}^{2} = 175 \left({x}^{2} - 2\right)$

This is clearly a quadratic, so make it equal to 0.

$75 \left({x}^{2} - 2\right) + 9 {\left({x}^{2} - 1\right)}^{2} - 175 \left({x}^{2} - 2\right) = 0$

Notice that the first and third terms are like terms, so we can add them together. Also square the middle term.

$9 \left({x}^{4} - 2 {x}^{2} + 1\right) - 100 \left({x}^{2} - 2\right) + = 0$

Remove the brackets by the distributive law:

$9 {x}^{4} - 18 {x}^{2} + 9 - 100 {x}^{2} + 200 = 0$

Simplify: $9 {x}^{4} - 118 {x}^{2} + 209 = 0$

Exploring the factors of 9 and 209 leads to
9 = 3x3, or 9x1 and 209 = 11 x 19

The combination of factors which adds to 118 is 99+19

Factorising gives $\left({x}^{2} - 11\right) \left(9 {x}^{2} - 19\right) = 0$

If ${x}^{2} - 11 = 0$
${x}^{2} = 11$
$x = \pm \sqrt{11}$

If $9 {x}^{2} - 19 = 0$
$9 {x}^{2} = 19$
${x}^{2} = \frac{19}{9}$

$x = \frac{\pm \sqrt{19}}{3}$