Solve for x:
a) 4x^2 + 3x - 2 = 0
b) 4x + cuberoot x - 2 = 0. (Hint: rename cuberoot x = z)
c)[(x + 4) / (x - 2)] - [(x - 3) / (x - 4)] = 0.
d) 3x = squareroot(3 + x)
This is a quadratic. If you don't know what else to do with it, use the quadratic formula. As this doesn't factor, that's what I'm going to do.
$\displaystyle 4x^2 + 3x - 2 = 0$
So....
$\displaystyle x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}$
$\displaystyle x = \frac{-3 \pm \sqrt{9 + 32}}{8}$
$\displaystyle x = \frac{-3 \pm \sqrt{41}}{8}$
-Dan
$\displaystyle 4x + \sqrt[3]{x} - 2 = 0$
Why were you given this problem? It seems above the level that I thought you were at.
Let $\displaystyle z = \sqrt[3]{x}$, then $\displaystyle x = z^3$.
Thus
$\displaystyle 4z^3 + z - 2 = 0$
This has no rational factors, so the only way to do this (neglecting Cardano's method, which I'll do if you really want me to) is to use a graphing calculator to estimate it, or numerically calculate it.
I get that
$\displaystyle z \approx 0.689398, -0.344699 \pm 0.778751 \cdot i$
-Dan
$\displaystyle \frac{x + 4}{x - 2} - \frac{x - 3}{x - 4} = 0$
First note that the solution set cannot contain 2 or 4, since that is outside the domain of the original problem.
Now add the fractions by getting a common denominator:
$\displaystyle \frac{x + 4}{x - 2} \cdot \frac{x - 4}{x - 4} - \frac{x - 3}{x - 4} \cdot \frac{x - 2}{x - 2} = 0$
$\displaystyle \frac{(x + 4)(x - 4) - (x - 3)(x - 2)}{(x - 2)(x - 4)} = 0$
$\displaystyle \frac{(x^2 - 16) - (x^2 - 5x + 6)}{(x - 2)(x - 4)} = 0$
$\displaystyle \frac{5x - 22}{(x - 2)(x - 4)} = 0$
The only way the LHS can be 0 is for the numerator be 0. Thus set
$\displaystyle 5x - 22 = 0$
$\displaystyle x = \frac{22}{5}$
(Note that x is neither 2 or 4, so this is a valid solution.)
-Dan
$\displaystyle 3x = \sqrt{3 + x}$
Square both sides:
$\displaystyle 9x^2 = 3 + x$
$\displaystyle 9x^2 - x - 3 = 0$
Again this is a quadratic that doesn't factor. So use the quadratic formula. I got:
$\displaystyle x = \frac{1 \pm \sqrt{109}}{18}$
Now, we need to check each of these in the original equation because squaring both sides tends to add spurious solutions. In this case we may simply note that if we use the "-" sign for x we don't have a solution. (I used the decimal approximation of the two x values to check in the original equation.)
So the solution is:
$\displaystyle x = \frac{1 + \sqrt{109}}{18}$
-Dan
$\displaystyle 4x + 3 \sqrt{x} - 2 = 0$
Let $\displaystyle z = \sqrt{x}$, that is to say let $\displaystyle x = z^2$.
Then
$\displaystyle 4z^2 + 3z - 2 = 0$
This is, in fact, the equation from part a), which I already solved. So we know that
$\displaystyle z = \frac{-3 \pm \sqrt{41}}{8}$
So
$\displaystyle x = z^2 = \left ( \frac{-3 \pm \sqrt{41}}{8} \right ) ^2$
$\displaystyle x = \frac{9 \mp 6\sqrt{41} + 41}{64} = \frac{50 \pm 6\sqrt{41}}{64}$ <-- We don't really care which order the + or - comes in.
$\displaystyle x = \frac{25 \pm 3 \sqrt{41}}{32}$
-Dan
Recall the quadratic formula.
For a quadratic equation of the form $\displaystyle ax^2 + bx + c = 0$
the roots of the equation are given by $\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$
For $\displaystyle 9x^2 - x - 3 = 0$, $\displaystyle a = 9$, $\displaystyle b = -1$ and $\displaystyle c = -3$
so under the square root we get: $\displaystyle b^2 - 4ac = (-1)^2 - (4)(9)(-3) = 109$