How do you simplify $\frac{3 m^{2} - 4 n^{2}}{2 m^{2} - n^{2}}$ $\frac { 3m ^ { 2} - 4n ^ { 2} } { 2m ^ { 2} - n ^ { 2} }$ ?

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Answer 1 · 2018-01-09T13:36:09

What we must use is the difference of two squares, using roots instead of square numbers.

From our normal factorising, we must first not the use of D.O.T.S

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2 = (a+b)(a-b)$

However, it must noted we don't have whole square numbers as coefficients, therefore these will simply remain as roots.

It should also be noted that $\sqrt{m^{2}} = m$ $sqrt(m^2) = m$
The same applies for $n$ $n$ , lets factorise and simplify.

$\frac{3 m^{2} - 4 n^{2}}{2 m^{2} - n^{2}}$ $(3m^2-4n^2)/(2m^2-n^2)$ can be simplified by factorising to:

$\frac{(\sqrt{3} m - 2 n) (\sqrt{3} m + 2 n)}{(\sqrt{2} m + n) (\sqrt{3} m - n)}$ $((sqrt(3)m-2n)(sqrt(3)m+2n))/((sqrt(2)m+n)(sqrt(3)m-n))$

This is the simplest form.

How do you simplify 3m2−4n22m2−n2\frac { 3m ^ { 2} - 4n ^ { 2} } { 2m ^ { 2} - n ^ { 2} } ?