Given that $a = \sqrt{\frac{b^2+4ac}{2a}}$, find the values of $b$ when $a = 4$ and $c = -1$. Answer b = ... Simplify $\left(\dfrac{125p^6}{q^3}\right)^{-1/3}\times\left(\dfrac{3p^0}{\sqrt{q}}\right)^2$, leaving your answer in positive index. Answer ... Solve the simultaneous equations. $2x - 3y = -15$ $0.6x + 1.5y = 6$ Answer $x =$ ... y = ... Express $\dfrac{6}{1-9x^2} + \dfrac{17}{6x^2+13x-5}$ as a single fraction in its simplest form. Answer ... Express $x^2 - 7x + 13.25$ in the form $(x - p)^2 + q$. Answer ... Hence, explain why the equation $k = x^2 - 7x + 13.25$ does not have solutions for some values of $k$. Answer ...

1. Substitute $a=4$ and $c=-1$ into the given formula.

$a=\sqrt{\dfrac{b^2+4ac}{2a}}$

When $a=4$ and $c=-1$,

$4=\sqrt{\dfrac{b^2+4(4)(-1)}{2(4)}}$

$4=\sqrt{\dfrac{b^2-16}{8}}$

2. Square both sides and solve the resulting equation for $b$.

$4^2=\dfrac{b^2-16}{8}$

$16=\dfrac{b^2-16}{8}$

$128=b^2-16$

$b^2=144$

$b=\pm 12$

1. Simplify the first factor using the negative fractional index.

$\left(\dfrac{125p^6}{q^3}\right)^{-1/3}=125^{-1/3}(p^6)^{-1/3}(q^{-3})^{-1/3}$
$=\dfrac{1}{5}p^{-2}q=\dfrac{q}{5p^2}$

2. Simplify the second factor and multiply the two factors, leaving only positive indices.

Since $p^0=1$ and $\sqrt{q}=q^{1/2}$,
$\left(\dfrac{3p^0}{\sqrt{q}}\right)^2=\left(\dfrac{3}{q^{1/2}}\right)^2=\dfrac{9}{q}$
Hence,
$\dfrac{q}{5p^2}\times\dfrac{9}{q}=\dfrac{9}{5p^2}$

1. Clear decimals in the second equation to make elimination easier.

$0.6x+1.5y=6$
Multiply by $10$:
$6x+15y=60$
Divide by $3$:
$2x+5y=20$

2. Eliminate $x$ by subtracting the first equation from the simplified second equation.

Given
$2x-3y=-15$
and
$2x+5y=20$
Subtract:
$(2x+5y)-(2x-3y)=20-(-15)$
$8y=35$
$y=\frac{35}{8}=4.375$

3. Substitute $y=\frac{35}{8}$ into one equation to find $x$.

Using $2x-3y=-15$,
$2x-3\left(\frac{35}{8}\right)=-15$
$2x-\frac{105}{8}=-15$
$2x=-15+\frac{105}{8}$
$2x=-\frac{120}{8}+\frac{105}{8}=-\frac{15}{8}$
$x=-\frac{15}{16}$

1. Factorise both denominators.

$1-9x^2=(1-3x)(1+3x)=-(3x-1)(3x+1)$
$6x^2+13x-5=(3x-1)(2x+5)$

2. Rewrite the expression using the factorised denominators.

$\frac{6}{1-9x^2}+\frac{17}{6x^2+13x-5}$
$=-\frac{6}{(3x-1)(3x+1)}+\frac{17}{(3x-1)(2x+5)}$

3. Use the lowest common denominator and combine the fractions.

Common denominator: $(3x-1)(3x+1)(2x+5)$.

$-\frac{6}{(3x-1)(3x+1)}+\frac{17}{(3x-1)(2x+5)}$
$=\frac{-6(2x+5)+17(3x+1)}{(3x-1)(3x+1)(2x+5)}$

4. Simplify the numerator and cancel the common factor.

$-6(2x+5)+17(3x+1)=-12x-30+51x+17=39x-13$
$39x-13=13(3x-1)$

So
$\frac{39x-13}{(3x-1)(3x+1)(2x+5)}=\frac{13(3x-1)}{(3x-1)(3x+1)(2x+5)}$
$=\frac{13}{(3x+1)(2x+5)}$

1. Complete the square for the quadratic and compensate for the added square term.

Half of $7$ is $3.5$, so
$x^2-7x+13.25=(x-3.5)^2-3.5^2+13.25.$

2. Simplify the constant term to obtain the required form.

$-3.5^2+13.25=-12.25+13.25=1.$
Hence,
$x^2-7x+13.25=(x-3.5)^2+1.$

1. Use the completed-square form to explain the range of possible values of the quadratic.

From part (e)(i),
$x^2-7x+13.25=(x-3.5)^2+1.$
Since $(x-3.5)^2\geq 0$ for all real $x$,
$x^2-7x+13.25\geq 1.$
Therefore $k=(x-3.5)^2+1$ has no real solution for $x$ when $k<1$.