# Reducible Equations

• Jul 14th 2010, 01:11 AM
jgv115
Reducible Equations
Solve:

a) $\displaystyle 4^{2x}- 5 .4 ^x +4 =0$

I know I have to make $\displaystyle u = 4x$ or something along those lines.

But I have no idea what to do next :(

Also,

Find the value of A, B, C if $\displaystyle A(x+1)^2 + Bx+C \equiv 2x^2 + 8x -6$

So I expand the first one...

$\displaystyle A(x^2+2x+1)+Bx+C$

then

$\displaystyle Ax^2 + 2Ax + A + Bx + C$

$\displaystyle Ax^2 + x(2A+B)+A+C$

With this I know that A = 2 B = -4 but I can't figure out C. I get -2 but it's wrong. What did I do wrong?
• Jul 14th 2010, 01:45 AM
Prove It
a) You actually let $\displaystyle u = 4^x$ so that you get a quadratic equation in $\displaystyle u$

$\displaystyle u^2 - 5u + 4 = 0$.

Solve it for $\displaystyle u$, then solve for $\displaystyle x$.

b) $\displaystyle A + C = -6$ and you know $\displaystyle A = 2$. Solve for $\displaystyle C$. I think you'll also find that $\displaystyle B = 4$, not $\displaystyle -4$.
• Jul 14th 2010, 01:54 AM
undefined
Oh I thought $\displaystyle 5.4^{x}$ meant $\displaystyle \left(\dfrac{54}{10}\right)^x$... I think it would be better to use \cdot to avoid confusion, as in $\displaystyle 5\cdot4^x$.
• Jul 14th 2010, 01:56 AM
Prove It
Quote:

Originally Posted by undefined
Oh I thought $\displaystyle 5.4^{x}$ meant $\displaystyle \left(\dfrac{54}{10}\right)^x$... I think it would be better to use \cdot to avoid confusion, as in $\displaystyle 5\cdot4^x$.

Or brackets, i.e. $\displaystyle 5\left(4^x\right)$.
• Jul 14th 2010, 02:33 AM
jgv115
ok, I get it now!

$\displaystyle 2+ C = -6$

So $\displaystyle C = -8$

Here is another (sorry I'm trying to pick this up myself with only the textbook and I'm trying my best :( )

Find the value of A,B,C if $\displaystyle Ax(x-1)+Bx^2+C(x-1)\equiv x-4-3x^2$

Expand: $\displaystyle Ax^2 - Ax+Bx^2+Cx-C$

I'm so sure what to do next so I group the terms into groups?

$\displaystyle x^2(A+B) -x(A-C) -C$

I'm pretty sure this is incorrect. Could someone help me out (not necessarily give me the answer)

Thanks for all the help!!!!
• Jul 14th 2010, 03:19 AM
Prove It
Quote:

Originally Posted by jgv115
ok, I get it now!

$\displaystyle 2+ C = -6$

So $\displaystyle C = -8$

Here is another (sorry I'm trying to pick this up myself with only the textbook and I'm trying my best :( )

Find the value of A,B,C if $\displaystyle Ax(x-1)+Bx^2+C(x-1)\equiv x-4-3x^2$

Expand: $\displaystyle Ax^2 - Ax+Bx^2+Cx-C$

I'm so sure what to do next so I group the terms into groups?

$\displaystyle x^2(A+B) -x(A-C) -C$

I'm pretty sure this is incorrect. Could someone help me out (not necessarily give me the answer)

Thanks for all the help!!!!

Technically what you have is right, but it's better to write it as

$\displaystyle (A + B)x^2 + (C-A)x - C \equiv -3x^2 + x - 4$.

That means

$\displaystyle A + B = -3$
$\displaystyle C - A = 1$
$\displaystyle -C = -4$.

Solve for $\displaystyle A, B, C$.
• Jul 14th 2010, 03:56 AM
jgv115
Thanks!! I finally get it!!!

I appreciate your help so much!! Thanks!!!!!!!!