Algebra — practice questions

A reliable method is to square both sides, as the modulus is always non-negative.
$(|x-3|)^2 \ge (|2x+1|)^2$
$(x-3)^2 \ge (2x+1)^2$
$x^2 - 6x + 9 \ge 4x^2 + 4x + 1$
$0 \ge 3x^2 + 10x - 8$
To find the roots of the quadratic $3x^2 + 10x - 8 = 0$, we can factorise it.
$(3x-2)(x+4) = 0$
So, the critical values are $x = \frac{2}{3}$ and $x = -4$.
The inequality is $3x^2 + 10x - 8 \le 0$. Since this is a positive parabola (opening upwards), the function is less than or equal to zero between its roots.
Therefore, the solution is $-4 \le x \le \frac{2}{3}$.

i) First, we must write the expression in the form $a(1+y)^n$.
$(4+x)^{1/2} = (4(1+\frac{x}{4}))^{1/2} = 4^{1/2}(1+\frac{x}{4})^{1/2} = 2(1+\frac{x}{4})^{1/2}$.
Now we apply the binomial formula with $n=\frac{1}{2}$ and 'x' replaced by '$\frac{x}{4}$'.
$2(1+\frac{x}{4})^{1/2} = 2 \left[ 1 + (\frac{1}{2})(\frac{x}{4}) + \frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(\frac{x}{4})^2 + \dots \right]$
$= 2 \left[ 1 + \frac{x}{8} + \frac{-\frac{1}{4}}{2}(\frac{x^2}{16}) + \dots \right]$
$= 2 \left[ 1 + \frac{x}{8} - \frac{x^2}{128} + \dots \right]$
$= 2 + \frac{x}{4} - \frac{x^2}{64} + \dots$\

ii) The expansion is valid when $|\frac{x}{4}| < 1$. This means $|x| < 4$, or $-4 < x < 4$.