If you note that the expressions on the left sides of the equations are perfect square trinomials, you can rewrite the expressions as follows:

\(\displaystyle x^{2}+4xy +4 y^{2} = 100\)

\(\displaystyle (x+2y)^{2} = 100\)

Either \(\displaystyle x+2y = 10\) or \(\displaystyle x+2y = -10\)

Similarly,

\(\displaystyle x^{2}-4xy +4 y^{2} = 64\)

\(\displaystyle (x-2y)^{2} = 64\)

Either \(\displaystyle x-2y= 8\) or \(\displaystyle x-2y = -8\)

Four systems of equations can be set up here, and in each case, we can find \(\displaystyle y\) by subtracting both sides:

\(\displaystyle x+2y = 10\)

\(\displaystyle \underline{x-2y= 8}\)

       \(\displaystyle 4y = 2\)

         \(\displaystyle y = \frac{1}{2}\)

\(\displaystyle x+2y =- 10\)

\(\displaystyle \underline{x-2y= -8}\)

       \(\displaystyle 4y = -2\)

         \(\displaystyle y =- \frac{1}{2}\)

\(\displaystyle x+2y = 10\)

\(\displaystyle \underline{x-2y= -8}\)

       \(\displaystyle 4y = 18\)

         \(\displaystyle y = \frac{9}{2}\)

\(\displaystyle x+2y =- 10\)

\(\displaystyle \underline{x-2y= 8}\)

       \(\displaystyle 4y = -18\)

         \(\displaystyle y =- \frac{9}{2}\)

The set of possible values of \(\displaystyle x\) is \(\displaystyle \left \{ -\frac{9}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{9}{2} \right \}\).

When we factor a polynomial, we are left with two factors of the form:

\(\displaystyle (x+a)(x+b)=0\)

For the given function, this means we must find two factors \(\displaystyle a\) and \(\displaystyle b\) whose product is 10 and whose sum is 7. Thinking about factors of 10, we can see that 1 and 10 cannot add up to 7 in any way. The only two factors left, then, are 2 and 5, which we can see have a product of 10 and a sum of 7. This gives us:

\(\displaystyle x^2+7x+10=0\)

\(\displaystyle (x+2)(x+5)=0\)

\(\displaystyle x=-2,x=-5\)

The roots of a function are the points at which it crosses the \(\displaystyle x\) axis, so \(\displaystyle f(x)\) will be equal to zero at these points. This means we set the function equal to zero, and then factor it to solve for the \(\displaystyle x\) values of the roots:

\(\displaystyle f(x)=2x^2+6x-20=0\)

\(\displaystyle 2(x^2+3x-10)=0\)

Now that we've factored out a \(\displaystyle 2\), we can see we need two factors whose product is \(\displaystyle -10\) and whose sum is \(\displaystyle 3\). Thinking about the possible factors, we can see that \(\displaystyle -2\) and \(\displaystyle 5\) have a product of \(\displaystyle -10\) and a sum of \(\displaystyle 3\). This gives us:

\(\displaystyle 2(x^2+3x-10)=0\)

\(\displaystyle 2(x+5)(x-2)=0\)

\(\displaystyle x=-5,x=2\)

In order to solve this problem, we need to combine like terms and then factor:

\(\displaystyle (x-3)^2=2x^2+14\)

\(\displaystyle x^2-6x+9=2x^2+14\)

\(\displaystyle 0=x^2+6x+5\)

\(\displaystyle 0=(x+5)(x+1)\)

\(\displaystyle x=-1, -5\)

To solve this we must factor:

The first step is to recognize that all of the terms on the equation's left side contain 4x. Therefore we can pull 4x out:

\(\displaystyle 4x(x^{2}-7x +12) = 0\)

Then we factor the inside of the parentheses, realizing that only -3, and -4 add to form -7, and multiply to form 12:

\(\displaystyle 4x(x - 3)(x - 4) = 0\)

Now we can see that, for the left side of the equation to equal zero, x can only equal 0, 3, or 4.

To solve, divide out a \(\displaystyle 2\) and then factor.

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

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

\(\displaystyle (x-4)(x+1)=0\)

\(\displaystyle x=4,-1\)

\(\displaystyle \frac{3x^{2}-27}{x-3}=\frac{3(x^{2}-9)}{x-3}=\frac{3(x-3)(x+3)}{x-3}\)

\(\displaystyle =3(x+3)=3x+9.\)

\(\displaystyle \left | 3x - 7 \right |=8\) really consists of two equations: \(\displaystyle 3x - 7 = \pm 8\)

We must solve them both to find two possible solutions.

\(\displaystyle 3x - 7 = 8 \Rightarrow 3x = 15\Rightarrow x = 5\)

\(\displaystyle 3x - 7 = - 8 \Rightarrow 3x = -1\Rightarrow x = -1/3\)

So \(\displaystyle x=5\) or \(\displaystyle x=\) \(\displaystyle \dpi{100} -\frac{1}{3}\).

It's actually easier to solve for the complement first.  Let's solve \(\displaystyle \left | 2x-5 \right |< 3\).  That gives -3 < 2x - 5 < 3.  Add 5 to get 2 < 2x < 8, and divide by 2 to get 1 < x < 4.  To find the real solution then, we take the opposites of the two inequality signs.  Then our answer becomes \(\displaystyle x\leq 1 \textsc{ or } x\geq 4\).

Set \(\displaystyle f (x) = 0\) and solve for \(\displaystyle x\):

\(\displaystyle f (x) = 0\)

\(\displaystyle \left | 4x + B \right | - 8 = 0\)

\(\displaystyle \left | 4x + B \right | = 8\)

Rewrite as a compound equation and solve each part separately:

\(\displaystyle 4x + B = -8 \textrm{ or } 4x + B = 8\)

\(\displaystyle 4x + B = -8\)

\(\displaystyle 4x + B -B = -B -8\)

\(\displaystyle 4x = -B -8\)

\(\displaystyle 4x \div 4 =\left ( -B -8 \right ) \div 4\)

\(\displaystyle x = \frac{-B -8 }{4}\)

\(\displaystyle 4x + B = 8\)

\(\displaystyle 4x + B -B = -B + 8\)

\(\displaystyle 4x = -B +8\)

\(\displaystyle 4x \div 4 =\left ( -B +8 \right ) \div 4\)

\(\displaystyle x = \frac{-B +8 }{4}\)