(i) Factorise each of the following integers:
(a) 713 (b) 1271 (c) 403
(ii) Solve the quadratic equation
$$ 403x^2 + 1271x + 713 = 0 $$
leaving your answer in surd form.
(iii) Simplify the following fractions:
$$ \frac{713}{1271}, \quad \frac{403}{1271}, \quad \frac{403}{713} $$
(i) (a) We have $\lceil \sqrt{713} \rceil = 27$, and $27^2 - 713 = 4^2$, and so $713 = (27-4)(27+4) = 23 \times 31$.
The factorisation is $713 = 23 \times 31$.
(b) $\lceil \sqrt{1271} \rceil = 36$, and $36^2 - 1271 = 5^2$, and so $1271 = (36-5)(36+5) = 31 \times 41$.
The factorisation is $1271 = 31 \times 41$
(c) $\lceil \sqrt{403} \rceil = 21$, and $21^2 - 403 = 38$, which is not a perfect square.
$22^2 - 403 = 9^2$, and so $403 = (22-9)(22+9) = 13 \times 31$.
The factorisation is $403 = 13 \times 31$
(ii) We can use the factorisations above
$$ \begin{align} 0 & = 403x^2 + 1271x + 713 \\ \\ & = (13 \cdot 31)x^2 + (31 \cdot 41)x + (23 \cdot 31) \\ \\ & = 13x^2 +41x +23 \end{align} $$
Using the general solution $x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we have
$$ \begin{align} x & = \frac{-41 \pm \sqrt{41^2-4(13)(23)}}{2(41)} \\ \\ x & = \frac{-41 \pm \sqrt{485}}{26} \end{align} $$
(iii) $$ \frac{713}{1271} = \frac{23 \cdot 31}{31 \cdot 41} = \frac{23}{41}$$
$$ \frac{403}{1271} = \frac{13 \cdot 31}{31 \cdot 41} = \frac{13}{41} $$
$$ \frac{403}{713} = \frac{13 \cdot 31}{23 \cdot 31} = \frac{13}{23} $$