Help would be greatly appreciated!
1) Solve : 2/3(x^2 + 3x - 1)^2 = 3/8
2) Solve the equation, by rearranging and completing the sqaure, finding the exact solutions - 6r^2 + 12r - pr^2 = 50
Thank you in advance!!
$\displaystyle \frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}$
$\displaystyle (x^2 + 3x - 1)^2 = \frac{9}{16}$
$\displaystyle x^2 + 3x - 1 = \pm \frac{3}{4}$
$\displaystyle x^2 + 3x + \left ( -1 \pm \frac{3}{4} \right ) = 0$
So you have two equations to solve:
$\displaystyle x^2 + 3x - \frac{1}{4} = 0$
and
$\displaystyle x^2 + 3x - \frac{7}{4} = 0$
-Dan
$\displaystyle -6r^2 + 12r - \pi r^2 = 50$
$\displaystyle -(6 + \pi )r^2 - 12r = 50$
Get the coefficient of the leading term to be 1:
$\displaystyle r^2 + \frac{12}{6 + \pi } \cdot r = -\frac{50}{6 + \pi }$
Now complete the square:
$\displaystyle r^2 + \frac{12}{6 + \pi } \cdot r + \frac{144}{4(6 + \pi)^2} = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}$
$\displaystyle \left ( r + \frac{12}{2(6 + \pi)} \right ) ^2 = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}$
I leave it to you to simplify and finish this out.
-Dan
1)
$\displaystyle \frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}$
$\displaystyle (x^2 + 3x - 1)^2 = \frac{3}{8} \times \frac{3}{2}$
$\displaystyle (x^2 + 3x - 1)^2 = \frac{9}{16}$
$\displaystyle x^2 + 3x - 1 = \frac{3}{4}$
$\displaystyle x^2 + 3x - 1 - \frac{3}{4} = 0$
$\displaystyle x^2 + 3x -\frac{7}{4}=0$
$\displaystyle (x - \frac{1}{2})(x + \frac{7}{2})$
$\displaystyle \implies x = \frac{1}{2} \ \mathrm{or} \ x = \frac{-7}{2}$
Haven't done question 2 as it's been done when I clicked Preview Post. Question 1 was done too but since I had done it too, I thought I might submit the reply.
The general idea for completing the square is this:
You have something of the form
$\displaystyle x^2 + 2ax = c$
(You can always reduce your problem to this form.)
Recall that $\displaystyle x^2 + 2ax + a^2 = (x + a)^2$
So we need to find a, then add $\displaystyle a^2$ to both sides.
In your case we have
$\displaystyle r^2 + \frac{12}{6 + \pi} \cdot r$
on the left hand side.
So we need to solve
$\displaystyle 2a = \frac{12}{6 + \pi} \implies a = \frac{12}{2(6 + \pi)}$
and then add
$\displaystyle a^2 = \frac{12^2}{2^2(6 + \pi)^2}$
to both sides.
-Dan