MaxWong

Q1. 

Rearranging the equation,

\(x^2 - 7x - 30 = 0\)

Then, using the quadratic formula, 

\(x = \dfrac{7 \pm \sqrt{7^2 - 4\cdot 1 \cdot (-30)}}{2}\)

I will leave the simplification for you.

Q2. Similar to Q1. If you understand what I did in Q1, then it's not hard. Just repeat what I did in Q1.

Q3. Keyword: "the ... equation ... has a double root."

The discriminant of a quadratic equation which has a double root is 0.

\(b^2 - 4\cdot 2 \cdot 18 = 0\\ b^2 - 12^2 = 0\\ (b - 12)(b + 12) = 0\)

Then this means b - 12 = 0 or b + 12 = 0. This is just one step from the answer, I will let you figure out this part.

Q4. This is a root relation problem. We will use the formulae for relations of roots.

\(\begin{cases} r + s = -\dfrac{(-6)}{1} = 6\\ rs = \dfrac{2}1 = 2 \end{cases}\)

We then express (r - s)^2 in terms of r + s and rs. 

\((r - s)^2 = r^2 - 2rs + s^2 = (r^2 + 2rs + s^2) - 4rs = (r + s)^2 -4rs\)

You can substitute the values of r + s and rs directly.

Q5. Similar to Q1. Repeat what I did.

Q6. 

\(\dfrac c{c - 5} = \dfrac4{c - 4}\)

Cross-multiplying,

\(c(c - 4) = 4(c- 5)\)

Expanding,

\(c^2 - 4c = 4c - 20\)

Moving everything to the left-hand side,

\(c^2 - 8c + 20 = 0\)

Then repeat what I did in Q1.

P.S. The entire "Quadratic Equation" part of the syllabus is for solving equations of the form ax^2 + bx + c = 0, so make sure you put the question in this form before you proceed.

P.P.S. Some parts need knowledge of discriminant \(\Delta = b^2 - 4ac\) and the root relation formulae. 

P.P.P.S. If you don't understand any part of my solutions, feel free to ask me.

Consider \((r + s)^2 = r^2 + s^2 + 2rs\).

\(r + s = -\dfrac43\\ rs = 4\)

So

\(\left(-\dfrac43\right)^2 = (r^2 + s^2) + 2\cdot 4\\ r^2 + s^2 = -\dfrac{56}9\)

\(90 \times 6 \times 4 = 540 \times 4 = \boxed{2160}\)

f(2) = 3

f(g(-1)) = 4(-1) + 7 = 3 = f(2)

Because f(g(-1)) = f(2), g(-1) = 2.

Use the fact that sum of interior angles of triangle add up to 180 degrees.

\(w + w + (4 w - 6) = 180\\ 6w = 186\\ w = 31\)

I will leave the other for you to finish.

Wow that's actually the answer

\(f^{-1}(3) = 4\\ f^{-1}(f^{-1}(3)) = f^{-1}(4) = 7\)

Let L be the side length of square.

By Law of Cosines,

\(L = \sqrt{1^2 + 1^2 - 2(1)(2)\cos 150^\circ} = \sqrt{2 + 2\sqrt 3}\)

Area of large square = \(L^2 = 2 + 2\sqrt 3\)

Substitute numbers,

\(\quad2f(2) - f(8) \\= 2(2^2 + 2\sqrt{2}) - (8^2 + 2\sqrt{8})\\=8 + 4\sqrt 2 - 64 - 4\sqrt 2\\ =-56\)

Did you directly copy it from somewhere else without proofreading over it?

Please do not, because the LaTeX may mess up.

Let length = L, width = W, height = H.

\(2(LW + WH + HL) = 100\\ L = 2W\\ L = H - 3\)

Changing the subjects of the second and third equations,

\(W = \dfrac L2\\ H = 3 + L\)

Substituting these into the first equation:

\(2\left(L \cdot \dfrac L2 + \dfrac L2 \cdot (3 + L) + (3 + L) \cdot L \right) = 100\\ L^2 + 3L(3 + L) = 100\\ 4L^2 + 9L - 100 = 0\)

Solving by quadratic formula gives \(L = 4\)

I believe you can work out the width and the height from here.