STEP I 2013 1

(i) We have:

y^2 + 3y - \dfrac 1 2 = 0

So that:

2y^2 + 6y - 1 = 0

This gives:

\displaystyle y = \frac {-6 \pm \sqrt {6^2 - 4 \times 2 \times (-1)}} 4 = \frac {-6 \pm \sqrt {44}} 4 = \frac {-3 \pm \sqrt {11}} 2

Note that 3 = \sqrt 9 < \sqrt 11, so the position root of this is:

\displaystyle y = \frac {-3 + \sqrt {11}} 2

So:

\displaystyle x = \frac 1 4 \left(-3 + \sqrt {11}\right)^2 = \frac 1 4 \left(9 - 6 \sqrt {11} + 11\right) = \frac 1 4 \left(20 - 6 \sqrt {11}\right) = \boxed{\frac 1 2 \left(10 - 3 \sqrt {11}\right)}

(ii)

(a)

As suggested by part (i), set \sqrt {x + 2} = u. Then u^2 = x + 2, so we write:

x + 10 \sqrt {x + 2} - 22 = (x + 2) + 10 \sqrt {x + 2} - 24 = u^2 + 10 u - 24 = 0

We see that:

u^2 + 10 u - 24 = (u + 12)(u - 2)

So u = 2 (since u \ge 0). Then \sqrt {x + 2} = 2, giving x + 2 = 4, so \boxed{x = 2}.

(b)

Write \sqrt {2x^2 - 8x - 3} = u. Then u^2 = 2x^2 - 8x - 3 so \dfrac 1 2 u^2 = x^2 - 4x - \dfrac 3 2. So write:

\displaystyle x^2 - 4x + \sqrt {2x^2 - 8x - 3} - 9 = \left(x^2 - 4x - \frac 3 2\right) + \sqrt {2x^2 - 8x - 3} - \frac {15} 2

So our equation is transformed into:

\displaystyle \frac 1 2 u^2 + u - \frac {15} 2 = 0

So that:

u^2 + 2u - 15 = (u - 3)(u + 5) = 0

So u = 3, ie. \sqrt {2x^2 - 8x - 3} = 3, so 2x^2 - 8x - 3 = 9, giving 2x^2 - 8x - 12 = 0, so x^2 - 4x - 6 = 0. We can write this as:

(x - 2)^2 - 10 = 0

So \boxed{x = -2 \pm \sqrt {10}}.