Solve the equation $x^2 - 6x - 23 = 0$ - Leaving Cert Mathematics - Question 4 - 2012

First, rearrange the first equation to express $s$ in terms of $r$: $s = 2r - 10$

Now, substitute this expression for $s$ into the second equation: $r(2r - 10) - (2r - 10)^2 = 12$

Expanding this gives: $2r^2 - 10r - (4r^2 - 40r + 100) = 12$

Rearranging terms yields: $-2r^2 + 30r - 100 - 12 = 0 \Rightarrow -2r^2 + 30r - 112 = 0$

Multiplying through by -1: $2r^2 - 30r + 112 = 0$

To solve for $r$, we can apply the quadratic formula again where:

• $a = 2, b = -30, c = 112$

Calculating the discriminant: $(-30)^2 - 4(2)(112) = 900 - 896 = 4$

Now applying the quadratic formula: $r = \frac{-(-30) \pm \sqrt{4}}{2(2)} = \frac{30 \pm 2}{4}$

This results in: $r = \frac{32}{4} = 8 ext{ or } r = \frac{28}{4} = 7$

For $r = 8$, substituting back into $s = 2(8) - 10$ yields: $s = 6$

For $r = 7$, substituting gives: $s = 2(7) - 10 = 4$

Thus the solutions are:

• For $r = 8$: $(r,s) = (8, 6)$
• For $r = 7$: $(r,s) = (7, 4)$