Expanding the bracket gives

\((x - 9)^2 = x^2 - 18x + 81\)

The \(7\) outside the bracket has no \(x\) in it, so it cannot change the \(x\) term. That makes \(b = -18\).

It is the completed-square rule read the other way round: the number in the bracket is half of \(b\), so \(b\) is double the number in the bracket.

The constant term on the right is made of two pieces: the \(81\) that comes out of squaring the bracket, and the \(7\) sitting outside it.

\(c = 81 + 7 = 88\)

The number inside the bracket is always half the coefficient of \(x\). Half of \(18\) is \(9\), so start with \((x + 9)^2\).

That bracket is not equal to the expression yet. Squaring it gives

\((x + 9)^2 = x^2 + 18x + 81\)

which has \(81\) on the end where the question has \(86\). Take the \(81\) away again and put the \(86\) back:

\(x^2 + 18x + 86 = (x + 9)^2 - 81 + 86 = (x + 9)^2 + 5\)

Every completed square works the same way: halve, square, subtract what you have added.

A quadratic written as \((x + p)^2 + q\) has its turning point at \((-p,\ q)\): the squared bracket is smallest when \(x = -p\), and the \(+q\) then sets how high that lowest point sits.

Here the turning point is \((7, -2)\), so \(-p = 7\) and \(q = -2\), which gives \(p = -7\).

\(y = (x - 7)^2 - 2\)

That is the same curve written a different way, so expanding it must give back \(x^2 + bx + c\):

\((x - 7)^2 = x^2 - 14x + 49\)

\(y = x^2 - 14x + 49 - 2 = x^2 - 14x + 47\)

\(b = -14\) and \(c = 47\)

Check it the other way round: half of \(-14\) is \(-7\), and \((-7)^2 = 49\), so the turning point of \(x^2 - 14x + 47\) is at \((7, -2)\) — which is where the question started.

Expand what the student wrote and compare it with what the question gave.

\((x - 8)^2 + 76 = x^2 - 16x + 64 + 76\)

That is not \(x^2 - 16x + 76\); it is \(64\) too big. Squaring the bracket manufactures \(64\) out of nothing, so \(64\) has to be taken away again before the constant is put back.

\(q = 76 - 64 = 12\)

Halve the coefficient of \(x\): half of \(-10\) is \(-5\).

\((x - 5)^2 = x^2 - 10x + 25\)

That bracket carries an extra \(25\) that the original does not have, so take it away again and put the real constant back:

\(x^2 - 10x + 26 = (x - 5)^2 - 25 + 26 = (x - 5)^2 + 1\)

Anything squared is either zero or positive — a square is never negative, whatever \(x\) is.

So \((x - 5)^2 \geq 0\), and adding \(1\) gives

\((x - 5)^2 + 1 \geq 1\)

The smallest value the expression can take is \(1\), which happens when \(x = 5\). Since \(1\) is greater than zero, the expression is always positive.

The word "hence" matters: it means use part (a). Trying a few values of \(x\) is not a proof, because it says nothing about the values not tried.

Half of \(-2\) is \(-1\), and \((x - 1)^2 = x^2 - 2x + 1\).

That is \(1\) on the end instead of \(21\), so subtract the \(1\) and add the \(21\):

\(x^2 - 2x + 21 = (x - 1)^2 - 1 + 21 = (x - 1)^2 + 20\)

In the form \(y = (x - 1)^2 + 20\), the bracket is squared, so it can never be negative. The smallest it can be is \(0\), and that happens when \(x - 1 = 0\), which is \(x = 1\).

At that value of \(x\) the whole of \(y\) is \(20\), and everywhere else \(y\) is bigger. So the turning point is \((1, 20)\), and because the \(x^2\) term is positive it is a minimum.

A parabola is symmetrical about the vertical line through its turning point, and part (b) put that point at \(x = 1\).

The line of symmetry is \(x = 1\).

Half of \(-12\) is \(-6\), and \((x - 6)^2 = x^2 - 12x + 36\), so \(x^2 - 12x\) on its own is \((x - 6)^2 - 36\).

Rewrite the equation with that in it:

\((x - 6)^2 - 36 + 29 = 0\)

\((x - 6)^2 = 7\)

Now undo the square. A positive number has two square roots, one of each sign, and dropping the negative one is where a mark is usually lost:

\(x - 6 = \pm\sqrt{ 7 }\)

\(x = 6 \pm \sqrt{ 7 }\)

\(7\) is not a square number, so \(\sqrt{ 7 }\) cannot be written as a whole number and this is the exact answer. Writing it as a decimal would lose accuracy, which is what "in surd form" is asking you not to do.

Completing the square only works on \(x^2\) with nothing in front of it, so take the \(4\) out of the first two terms — and only those two:

\(4x^2 + 32x + 87 = 4(x^2 + 8x) + 87\)

Inside the bracket, half of \(8\) is \(4\), so

\(x^2 + 8x = (x + 4)^2 - 16\)

Put that back in:

\(4((x + 4)^2 - 16) + 87\)

The \(4\) outside still has to multiply both pieces inside, and \(4 \times 16 = 64\):

\(4(x + 4)^2 - 64 + 87 = 4(x + 4)^2 + 23\)

The step people lose marks on is the last one: the number taken out of the bracket has to be multiplied back in before it is combined with \(87\).

The method does not change because the numbers are odd: the number inside the bracket is still half the coefficient of \(x\).

Half of \(11\) is \(\frac{11}{2}\), so start with \((x + \frac{11}{2})^2\).

Squaring that bracket gives

\((x + \frac{11}{2})^2 = x^2 + 11x + \frac{121}{4}\)

which carries \(\frac{121}{4}\) on the end where the expression has \(-10\). Take the \(\frac{121}{4}\) away and put the \(-10\) back:

\(x^2 + 11x - 10 = (x + \frac{11}{2})^2 - \frac{121}{4} - 10 = (x + \frac{11}{2})^2 - \frac{161}{4}\)

The quarters are not a mistake. Squaring a half always makes a quarter, so an odd coefficient of \(x\) always leaves a quarter behind.

An expression written as \((x + p)^2 + q\) has minimum value \(q\), because a square is never negative and is zero when \(x = -p\). So complete both squares and compare the two constants.

For \(A\), half of \(-10\) is \(-5\):

\(x^2 - 10x + 12 = (x - 5)^2 - 25 + 12 = (x - 5)^2 - 13\)

so the minimum value of \(A\) is \(-13\).

For \(B\), half of \(8\) is \(4\):

\(x^2 + 8x + 15 = (x + 4)^2 - 16 + 15 = (x + 4)^2 - 1\)

so the minimum value of \(B\) is \(-1\).

\(-13\) is less than \(-1\), so A has the smaller minimum value. It reaches it at \(x = 5\).

Comparing the two constant terms \(12\) and \(15\) is not a method, and here it happens to point the same way — but only by luck. Each \(x\) term slides its own curve down by a different amount, and only the completed square says by how much.