Reducible Equations

**jgv115** · July 14th 2010, 01:11 AM

Solve:

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

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

But I have no idea what to do next

Also,

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

So I expand the first one...

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

then

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

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

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?

**Prove It** · July 14th 2010, 01:45 AM

a) You actually let $u = 4^x$ so that you get a quadratic equation in $u$

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

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

b) $A + C = -6$ and you know $A = 2$ . Solve for $C$ . I think you'll also find that $B = 4$ , not $-4$ .

**undefined** · July 14th 2010, 01:54 AM

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

**Prove It** · July 14th 2010, 01:56 AM

Originally Posted by undefined

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

Or brackets, i.e. $5\left(4^x\right)$ .

**jgv115** · July 14th 2010, 02:33 AM

ok, I get it now!

$2+ C = -6$

So $C = -8$

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 $Ax(x-1)+Bx^2+C(x-1)\equiv x-4-3x^2$

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

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

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

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

Thanks for all the help!!!!

**Prove It** · July 14th 2010, 03:19 AM

Originally Posted by jgv115

ok, I get it now!

$2+ C = -6$

So $C = -8$

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 $Ax(x-1)+Bx^2+C(x-1)\equiv x-4-3x^2$

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

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

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

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

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

That means

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

Solve for $A, B, C$ .

**jgv115** · July 14th 2010, 03:56 AM

Thanks!! I finally get it!!!

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

Thread: Reducible Equations

LinkBack

Thread Tools

Display

Reducible Equations

Similar Math Help Forum Discussions

Reducible

Reducible Polynomials

reducible and irreducible

R[x] - Reducible?

Equations Reducible To Quadratics