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

2. Originally Posted by steph_r
Help would be greatly appreciated!

1) Solve : 2/3(x^2 + 3x - 1)^2 = 3/8
$\frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}$

$(x^2 + 3x - 1)^2 = \frac{9}{16}$

$x^2 + 3x - 1 = \pm \frac{3}{4}$

$x^2 + 3x + \left ( -1 \pm \frac{3}{4} \right ) = 0$

So you have two equations to solve:
$x^2 + 3x - \frac{1}{4} = 0$
and
$x^2 + 3x - \frac{7}{4} = 0$

-Dan

3. Originally Posted by steph_r
2) Solve the equation, by rearranging and completing the sqaure, finding the exact solutions - 6r^2 + 12r - pr^2 = 50
$-6r^2 + 12r - \pi r^2 = 50$

$-(6 + \pi )r^2 - 12r = 50$

Get the coefficient of the leading term to be 1:
$r^2 + \frac{12}{6 + \pi } \cdot r = -\frac{50}{6 + \pi }$

Now complete the square:
$r^2 + \frac{12}{6 + \pi } \cdot r + \frac{144}{4(6 + \pi)^2} = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}$

$\left ( r + \frac{12}{2(6 + \pi)} \right ) ^2 = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}$

I leave it to you to simplify and finish this out.

-Dan

4. Originally Posted by steph_r
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

1)
$\frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}$
$(x^2 + 3x - 1)^2 = \frac{3}{8} \times \frac{3}{2}$
$(x^2 + 3x - 1)^2 = \frac{9}{16}$
$x^2 + 3x - 1 = \frac{3}{4}$
$x^2 + 3x - 1 - \frac{3}{4} = 0$
$x^2 + 3x -\frac{7}{4}=0$
$(x - \frac{1}{2})(x + \frac{7}{2})$
$\implies x = \frac{1}{2} \ \mathrm{or} \ x = \frac{-7}{2}$

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.

5. Originally Posted by Air
1)
$\frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}$
$(x^2 + 3x - 1)^2 = \frac{3}{8} \times \frac{3}{2}$
$(x^2 + 3x - 1)^2 = \frac{9}{16}$
$x^2 + 3x - 1 = \frac{3}{4}$
$x^2 + 3x - 1 - \frac{3}{4} = 0$
$x^2 + 3x -\frac{7}{4}=0$
$(x - \frac{1}{2})(x + \frac{7}{2})$
$\implies x = \frac{1}{2} \ \mathrm{or} \ x = \frac{-7}{2}$

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.
Thank you very much!! I really appreciate it.

6. ## Completing the square - extra help

...

Originally Posted by topsquark
$-6r^2 + 12r - \pi r^2 = 50$

$-(6 + \pi )r^2 - 12r = 50$

Get the coefficient of the leading term to be 1:
$r^2 + \frac{12}{6 + \pi } \cdot r = -\frac{50}{6 + \pi }$

Now complete the square:
$r^2 + \frac{12}{6 + \pi } \cdot r + \frac{144}{4(6 + \pi)^2} = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}$

Sorry, but may i ask how you went about the completing the square above? I dont quite understand

$\left ( r + \frac{12}{2(6 + \pi)} \right ) ^2 = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}$

I leave it to you to simplify and finish this out.

-Dan

7. The general idea for completing the square is this:

You have something of the form
$x^2 + 2ax = c$
(You can always reduce your problem to this form.)

Recall that $x^2 + 2ax + a^2 = (x + a)^2$

So we need to find a, then add $a^2$ to both sides.

$r^2 + \frac{12}{6 + \pi} \cdot r$
on the left hand side.

So we need to solve
$2a = \frac{12}{6 + \pi} \implies a = \frac{12}{2(6 + \pi)}$
$a^2 = \frac{12^2}{2^2(6 + \pi)^2}$
to both sides.

-Dan

8. ## Thank you!

..

Originally Posted by topsquark
The general idea for completing the square is this:

You have something of the form
$x^2 + 2ax = c$
(You can always reduce your problem to this form.)

Recall that $x^2 + 2ax + a^2 = (x + a)^2$

So we need to find a, then add $a^2$ to both sides.

$r^2 + \frac{12}{6 + \pi} \cdot r$
on the left hand side.

So we need to solve
$2a = \frac{12}{6 + \pi} \implies a = \frac{12}{2(6 + \pi)}$
$a^2 = \frac{12^2}{2^2(6 + \pi)^2}$