The quickest route is to factorise once and then substitute, because the factorised form makes the sign obvious without any squaring.

\(x^2 - 11x + 28 = (x - 4)(x - 7)\)

Put \(x = 6\) into that:

\((6 - 4)(6 - 7) = 2 \times (-1) = -2\)

\(-2\) is less than zero, so the answer is Yes.

What is really going on: the expression is negative exactly between the two roots \(4\) and \(7\), and \(6\) either is or is not in that stretch. Substituting is the way to show it rather than assert it.

\(f(x) > 0\) asks where the curve is above the \(x\)-axis.

The sketch is inverted (n-shaped), so the curve is above the axis between the two crossings.

The crossings are \(x = -2.5\) and \(x = -1\), so

\(-2.5 < x < -1\)

The numbers whose square is exactly 25 are \(5\) and \(-5\), because \(5^2 = 25\) and \((-5)^2 = 25\).

A square is large only when x is far from zero on either side, so the solution is two separate pieces, one beyond each root.

\(x \lt -5\) or \(x \gt 5\)

Check with \(x = 0\), which gives \(0^2 = 0\). The statement \(0 \gt 25\) is false, so \(0\) is not in the solution, as the answer above shows.

The expression on the left is the \(y\) of the curve, so the question is asking which values of \(x\) put the curve below the \(x\)-axis.

Read the two crossings off the grid: they are at \(x = 1\) and \(x = 5\).

Between those two the curve dips under the axis, so

\(1 < x < 5\)

This time the curve has to be above the \(x\)-axis, which is everywhere outside the two crossings — the left arm and the right arm.

\(x < 1\) or \(x > 5\)

The two answers together account for every value of \(x\) except \(1\) and \(5\) themselves, where the curve is exactly on the axis and neither inequality holds.

Get \(x^2\) on its own first by dividing both sides by \(4\). Dividing by a positive number leaves the inequality sign alone.

\(x^2 \gt 16\)

Now think about which numbers have a square that big. Squaring throws away the sign, so it is the distance from zero that has to be more than \(4\), and there are two ways to be that far out.

\(x \gt 4\) — for instance \(5^2 = 25\), which is more than \(16\).

\(x \lt -4\) — and \((-5)^2 = 25\) is more than \(16\) as well.

\(x \lt -4\) or \(x \gt 4\)

The answer is two separate pieces, not one stretch. Writing \(-4 \leq x \leq 4\) would be the solution of the opposite inequality — those are exactly the values whose square is not big enough.

Squaring loses the sign, so \(x^2 = 49\) has two roots, \(x = 7\) and \(x = -7\). Solve \(x^2 \gt 49\) for yourself, then compare with the drawing.

The values that work are the ones outside \(-7\) and \(7\).

Test a value where the drawing differs from that. Take \(x = 8\), which is not shaded on Yusuf's line.

\(8^2 = 64\), which is more than 49, so \(x = 8\) should be shaded.

The line is wrong at \(x = 8\).

Start by finding where the expression is exactly zero, because those are the only places it can change sign.

\(x^2 -x - 6 = 0\)

\((x + 2)(x - 3) = 0\)

\(x = -2\) or \(x = 3\)

Now think about the shape. The \(x^2\) term is positive, so the graph of \(y = x^2 -x - 6\) is a U, crossing the \(x\)-axis at \(-2\) and at \(3\). A U-shaped curve dips below the axis between its two crossings and is above it outside them.

The question asks where the expression is below zero, so it is the middle piece:

\(-2 < x < 3\)

Both ends are strict, because at \(x = -2\) and \(x = 3\) the expression is equal to zero, not less than it.

Find the critical values first — the places where the expression is zero, which are the only places its sign can change.

\((x + 5)(x + 3) = 0\) gives \(x = -5\) and \(x = -3\).

Those two numbers cut the number line into three stretches. Test one value from each and see what sign comes out.

At \(x = -6\): \((-6)^2 + 8(-6) + 15 = 3\), which is positive.

At \(x = -4\): \((-4)^2 + 8(-4) + 15 = -1\), which is negative.

At \(x = -2\): \((-2)^2 + 8(-2) + 15 = 3\), which is positive.

The two outer stretches are the ones that work. The sign is \(\geq\) rather than \(>\), so the critical values themselves are included — at \(x = -5\) and \(x = -3\) the expression is zero, and zero does satisfy "greater than or equal to zero".

\(x \leq -5\) or \(x \geq -3\)

Two separate stretches cannot be written as one chain: \(-3 \leq x \leq -5\) would say \(x\) is bigger than \(-3\) and smaller than \(-5\) at the same time, which no number is.

Solve the matching equation first to find where the expression is zero:

\((x + 7)(x + 1) = 0\) gives \(x = -7\) and \(x = -1\).

The \(x^2\) term is positive, so the curve is a U and is at or below the axis between the two crossings, ends included:

\(-7 \leq x \leq -1\)

Now write down every whole number in that stretch. The sign is \(\leq\), so \(-7\) and \(-1\) are both in the list:

\(-7,\ -6,\ -5,\ -4,\ -3,\ -2,\ -1\)

That is 7 integers. Counting them is a quick check — from \(-7\) to \(-1\) inclusive there should be \(-1 - (-7) + 1 = 7\) of them.

Nothing can be factorised until one side is zero, so move every term to the left:

\(x^2 + 5x - 14 > 0\)

Take care with the signs — every term that crosses the \(>\) changes sign, and the \(>\) itself does not change, because moving terms is adding and subtracting, not multiplying by a negative.

Now find the critical values:

\((x + 7)(x - 2) = 0\) gives \(x = -7\) and \(x = 2\).

The curve \(y = x^2 + 5x - 14\) is a U crossing the axis at those two points, and the question wants it strictly above the axis, which is the two arms rather than the dip between them:

\(x < -7\) or \(x > 2\)