MaxWong

$a = \log_{16} 49 = \dfrac12 \log_4 49 =\log_4 7$

$ab = \log_47 \cdot \log_725 = \dfrac{\ln 7}{\ln 4}\cdot \dfrac{\ln25}{\ln7} = \dfrac{\ln25}{\ln4} = \log_4 25$

$ab + 1 = \log_4 25 + 1 = \log_425 + \log_44 = \log_4100$

That means $4^{ab + 1} = 4^{\log_4 100} = 100$

To find the x-intercepts, we let y = 0 and solve for x. If x = a is a possible value of x, that corresponds to the x-intercept (a, 0).

But before that, we need to simplify it first.

$y = 2x^2 +13x - 7 - x^2 - 6x \\y= x^2 + 7x - 7$

Then let y = 0, and solve the equation using quadratic formula:

$x^2 + 7x - 7 = 0\\ x= \dfrac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}\\ x = \boxed{\phantom{\dfrac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}}}$

Each solution corresponds to one x-intercept, as I explained above.

Using the sum of G.S. formula, we have 

$\dfrac{\dfrac b7}{1 - \dfrac b7} = 7\\ \dfrac b7 = 7 - b\\ \dfrac {8b}7 = 7\\ b = \dfrac{49}8$

I don't think there is enough information to calculate the length of EC. If someone comes up with a solution, I am more than happy to see it.

hipie: 

In fact, complex square root (in mathematical terms: $\sqrt{\phantom{a}} : \mathbb C \to \mathbb C$) is multi-valued, which means it can have multiple outputs for one input. Moreover, for any non-zero complex number, there are always two distinct complex square roots. Therefore, it is actually correct to say that $\sqrt 9 = \pm 3$. But generally, if it is not specified and the expression inside the square root is real, then we assume that $\sqrt{\phantom{a}}$ means real-valued square root.

Hint: 100x^2 + 20x + 1 = (10x + 1)^2.

If (10x + 1)^2 = 0, then 10x + 1 = 0. The answer is quite obvious now.

Simplify this a little:

f(x) = 3x^2 - 4x - x^2 + 7x + 8 = 2x^2 + 3x + 8

This is a quadratic function, which means its graph is a parabola. The axis of symmetry of the graph is $x = - \dfrac{3}{2(2)} = -\dfrac34$. (It is always x = -b/(2a) for the parabola y = ax^2 + bx + c.)

That means $f(x) = f\left(2\left(-\dfrac34\right) - x\right) = f\left(-\dfrac32 - x\right)$. The required constant is -3/2.

Converting back to multiplication gives:

$\quad \dfrac{\sqrt{49x}}{\sqrt{21y}} \div \dfrac{\sqrt{8z}}{\sqrt{xyz}} \div \dfrac{\sqrt{338x}}{\sqrt{18y}}\\ =\dfrac{\sqrt{49x}}{\sqrt{21y}} \times \dfrac{\sqrt{xyz}}{\sqrt{8z}} \times \dfrac{\sqrt{18y}}{\sqrt{338x}}\\ = \dfrac{21\sqrt 2 xy \sqrt z}{52 \sqrt{21}\sqrt{xyz}}\\ = \dfrac{\sqrt{42xy}}{52}$

(a) By Cauchy-Schwarz inequality, 

$(c^4 + d^4)(e^4 + f^4) \geq ((ce)^2 + (df)^2)^2$

Then, again, by Cauchy-Schwarz inequality,

$\begin{array}{rcl} (a^2 + b^2)^2(c^4 + d^4)(e^4 + f^4) &\geq & (a^2 + b^2)^2((ce)^2 + (df)^2)^2\\ &=& \left((a^2 + b^2)\left((ce)^2+ (df)^2\right)\right)^2\\ &\geq& ((ace + bdf)^2)^2\\ &=& (ace + bdf)^4 \end{array}$

For (b), you can instead prove that $(a^2 + b^2)^2 (c^2 + d^2)^2 (e^2 + f^2)^2 \geq (ace + bdf)^4$. The rest follows from the fact that $(x + y)^2 \geq x^2 + y^2$ for nonnegative x, y, and part (a). If you can prove this inequality, the result of part (b) immediately follows.

The method hipie gave is also possible, but it is tedious. Instead, we could simplify the square roots on their own first.

$\quad \dfrac{\sqrt{160}}{\sqrt{252}} \times \dfrac{\sqrt{245}}{\sqrt 3}\\ =\dfrac{4 \sqrt{10}}{6 \sqrt 7} \times \dfrac{7 \sqrt 5}{\sqrt 3}\\ = \dfrac{4\times 7}{6} \times \sqrt{\dfrac{10 \times 5}{7 \times 3}}\\ = \dfrac{4\times 7 \times \sqrt{10 \times 5 \times 7 \times 3}}{6\times 7 \times 3}\\ = \dfrac{2\sqrt{1050}}{9}\\ = \dfrac{10 \sqrt{42}}{9}$