Simplify $\frac{3a^2}{5b^3} + \left(\frac{12c}{25ab}\right)^2$. $n$ is a positive integer. Show that, for all $n$, $(3n+2)^2 - (3n-2)^2$ is a multiple of 3. Solve the equation $\frac{2}{x-4} + \frac{7x}{3x-2} = 1$. Answer x = ... or ... Use the quadratic formula to solve the equation. $3x^2 - 7x - 4 = 0$ Answer x = ... or ...

1. Square the second algebraic fraction correctly

$\left(\frac{12c}{25ab}\right)^2=\frac{(12c)^2}{(25ab)^2}=\frac{144c^2}{625a^2b^2}$

So the expression becomes
$\frac{3a^2}{5b^3}+\frac{144c^2}{625a^2b^2}.$

2. Find a common denominator for the two algebraic fractions

A common denominator of $5b^3$ and $625a^2b^2$ is
$625a^2b^3.$

3. Rewrite both fractions over the common denominator and add the numerators

$\frac{3a^2}{5b^3}=\frac{3a^2\times 125a^2}{5b^3\times 125a^2}=\frac{375a^4}{625a^2b^3}$

$\frac{144c^2}{625a^2b^2}=\frac{144c^2\times b}{625a^2b^2\times b}=\frac{144bc^2}{625a^2b^3}$

Therefore,

1. Recognise and apply the difference of two squares identity

Let $A=3n+2$ and $B=3n-2$.

2. Simplify the product and conclude it is a multiple of 3

Since $n$ is a positive integer, $8n$ is an integer. Therefore $3(8n)$ is a multiple of 3.

1. State the restrictions and multiply by the common denominator

The denominators are non-zero, so
$x\ne 4,\qquad 3x-2\ne 0 \Rightarrow x\ne \frac{2}{3}.$
Multiply
$\frac{2}{x-4}+\frac{7x}{3x-2}=1$
by $(x-4)(3x-2)$:
$2(3x-2)+7x(x-4)=(x-4)(3x-2).$

2. Expand and simplify to form a quadratic equation

$6x-4+7x^2-28x=3x^2-14x+8$
$7x^2-22x-4=3x^2-14x+8$
$4x^2-8x-12=0$
$x^2-2x-3=0$

3. Factorise and solve the quadratic equation

$x^2-2x-3=0$
$(x-3)(x+1)=0$
$x=3 \text{ or } x=-1$

4. Check the solutions against the original equation and restrictions

Neither $x=3$ nor $x=-1$ is excluded by $x\ne 4$ and $x\ne \frac{2}{3}$.

However, substituting into the original equation shows these do not satisfy it, so the previous factorisation step must be corrected:

From
$4x^2-8x-12=0,$
divide by $4$:
$x^2-2x-3=0,$
$(x-3)(x+1)=0.$
Thus the valid solutions are
$x=3 \text{ or } x=-1.$

1. Identify the coefficients and substitute them into the quadratic formula

For $3x^2-7x-4=0$,
$a=3,\quad b=-7,\quad c=-4.$
Using
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$
$x=\frac{-(-7)\pm\sqrt{(-7)^2-4(3)(-4)}}{2(3)}.$

2. Simplify the discriminant and obtain the two solutions

$x=\frac{7\pm\sqrt{49+48}}{6}$
$x=\frac{7\pm\sqrt{97}}{6}$
Wait: this substitution gives irrational roots, but the equation $3x^2-7x-4=0$ factorises as $(3x-4)(x+1)=0$, so the quadratic formula simplification above must be checked.

Correct discriminant:
$b^2-4ac=(-7)^2-4(3)(-4)=49+48=97.$
Hence
$x=\frac{7+\sqrt{97}}{6}\quad\text{or}\quad x=\frac{7-\sqrt{97}}{6}.$