Write down the three numbers before anything else — getting a sign wrong here is the commonest way to lose all three marks.

\(a = 2\), \(b = 9\), \(c = -9\)

The part under the root first:

\(b^2 - 4ac = (9)^2 - 4 \times 2 \times (-9) = 81 - (-72) = 153\)

\(153\) is not a square number, so this quadratic has no whole-number factorisation and the formula is the way through.

\(x = \frac{ -9 \pm \sqrt{ 153 } }{ 4 }\)

\(\sqrt{ 153 } = 12.369\) to 3 decimal places, so

\(x = \frac{ -9 + 12.369 }{ 4 } = 0.84\)

\(x = \frac{ -9 - 12.369 }{ 4 } = -5.34\)

Carrying an extra decimal place through \(\sqrt{ 153 }\) is what stops the rounding creeping into the second decimal place of the answers.

The equation is already in the form the formula wants, with

\(a = 1\), \(b = 1\), \(c = -47\)

\(b^2 - 4ac = 1 - 4 \times 1 \times (-47) = 1 + 188 = 189\)

Taking away a negative is what makes the discriminant so large here, and it is also what guarantees the two roots have opposite signs.

\(x = \frac{ -1 \pm \sqrt{ 189 } }{2}\), and \(\sqrt{ 189 } = 13.748\) to 3 decimal places.

\(x = \frac{ -1 + 13.748 }{2} = 6.37\)

\(x = \frac{ -1 - 13.748 }{2} = -7.37\)

The second root is negative, and a width cannot be negative, so \(x = -7.37\) is rejected.

The width is 6.37 cm.

Worth a check: the length is then about 7.37 cm, and \(6.37 \times 7.37\) comes to about 47.

An equation has exactly one solution when the discriminant \(b^2 - 4ac\) is zero.

Here \(a = 1\) and \(b = 4\), so

\((4)^2 - 4 \times 1 \times c = 0\)

\(16 = 4c\)

\(c = 16 \div 4 = 4\)

Check: \(x^2 + 4x + 4 = (x + 2)^2\), a perfect square, which is exactly the case where the two solutions coincide.

In the quadratic formula the \(\pm\) is what produces two solutions. An equation has exactly one solution when the part under the root is zero, so

\(b^2 - 4ac = 0\)

Here \(a = 2\), \(b = k\) and \(c = 162\), which gives

\(k^2 - 4 \times 2 \times 162 = 0\)

\(k^2 = 1296\)

A number squared is \(1296\) in two ways, one positive and one negative:

\(k = 36\) or \(k = -36\)

Both are real answers. Writing only \(36\) throws away the equation \(2x^2 - 36x + 162 = 0\), which has the repeated solution too.

\(a = 1\), \(b = -3\), \(c = -1\)

\(b^2 - 4ac = (-3)^2 - 4 \times 1 \times (-1) = 9 - (-4) = 13\)

Putting that into the formula:

\(x = \frac{ 3 \pm \sqrt{ 13 } }{2}\)

Now stop. \(13\) has no square factor, so \(\sqrt{ 13 }\) cannot be written any more simply, and \(3\) is odd, so nothing cancels with the \(2\) underneath either. The fraction is already as tidy as it goes.

"In surd form" means leaving the root in the answer. Turning it into a decimal here would throw away accuracy for nothing, and would not earn the marks.

\(a = 1\), \(b = -9\), \(c = -2\)

\(b^2 - 4ac = (-9)^2 - 4 \times 1 \times (-2) = 81 - (-8) = 89\)

\(b^2 - 4ac\) sits under the square root in the quadratic formula, so its sign decides what the formula can produce.

Here it is \(89\), which is positive. A positive number has two square roots, one of each sign, so the plus-or-minus in the formula gives two different answers.

So the equation has two real solutions.

The formula only works on an equation with a zero on one side, so move the \(16\) across first:

\(3x^2 - 4x - 16 = 0\)

Now read off the three numbers. The one that catches people is \(c\): it is \(-16\), not \(16\).

\(a = 3\), \(b = -4\), \(c = -16\)

\(b^2 - 4ac = (-4)^2 - 4 \times 3 \times (-16) = 16 - (-192) = 208\)

\(x = \frac{ 4 \pm \sqrt{ 208 } }{ 6 }\), and \(\sqrt{ 208 } = 14.422\) to 3 decimal places.

\(x = \frac{ 4 + 14.422 }{ 6 } = 3.07\)

\(x = \frac{ 4 - 14.422 }{ 6 } = -1.74\)

Put \(h = 18\) into the formula and take everything to one side, because the quadratic formula only works on an equation equal to zero.

\(22t - 5t^2 = 18\)

\(5t^2 - 22t + 18 = 0\)

Turning it round so the \(t^2\) term is positive is worth the extra line. The formula works either way, but the signs are easier to keep straight.

\(a = 5\), \(b = -22\), \(c = 18\)

\(b^2 - 4ac = (-22)^2 - 4 \times 5 \times 18 = 484 - 360 = 124\)

\(t = \frac{ 22 \pm \sqrt{ 124 } }{ 10 }\)

\(\sqrt{ 124 } = 11.136\) to 3 decimal places, so

\(t = \frac{ 22 - 11.136 }{ 10 } = 1.09\)

\(t = \frac{ 22 + 11.136 }{ 10 } = 3.31\)

Both times are answers, and that is the whole point of the question. The ball passes \(18\) m on the way up after \(1.09\) seconds, keeps rising until \(2.2\) seconds, and passes \(18\) m again on the way back down after \(3.31\) seconds. Throwing the smaller root away — which is the right move when the answer is a length — would lose a mark here.

There is an \(x\) underneath, so multiply every term by \(x\) to clear it.

\(x^2 -x = 27\)

Take the \(27\) across so that one side is zero — the formula only works on an equation in that shape.

\(x^2 -x - 27 = 0\)

\(a = 1\), \(b = -1\), \(c = -27\)

\(b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-27) = 1 + 108 = 109\)

\(x = \frac{ 1 \pm \sqrt{ 109 } }{ 2 }\)

\(\sqrt{ 109 } = 10.440\) to 3 decimal places, so

\(x = \frac{ 1 + 10.440 }{ 2 } = 5.72\)

\(x = \frac{ 1 - 10.440 }{ 2 } = -4.72\)

Both answers count. \(x = 0\) is the only value the original equation could not have, and neither of these is zero.

A border of width \(x\) adds \(x\) at both ends of each side, so the whole mount measures \((21 + 2x)\) cm by \((11 + 2x)\) cm.

\((21 + 2x)(11 + 2x) = 362\)

Multiply the brackets out:

\(231 + 42x + 22x + 4x^2 = 362\)

\(4x^2 + 64x + 231 = 362\)

\(4x^2 + 64x - 131 = 0\)

Now the formula, with \(a = 4\), \(b = 64\) and \(c = -131\):

\(b^2 - 4ac = (64)^2 - 4 \times 4 \times (-131) = 4096 + 2096 = 6192\)

\(\sqrt{ 6192 } = 78.689\) to 3 decimal places, so

\(x = \frac{ -64 \pm 78.689 }{ 8 }\)

The minus sign gives \(x = -17.84\), and a border cannot have a negative width, so that root is thrown away.

\(x = 1.84\)

Where the \(131\) came from is worth seeing: the photograph itself covers \(231\) cm², and \(231 + 131 = 362\) cm², which is the area the question gave for the two together.