quadratic equations: solving x^-2x-13=0 using method

[bbayes81]

I need some help on a series of problems. This is part of a project. I am to use the following method for solving a quadratic equation and I am getting very confused when I get to part (d). I will write out the entire method, and I would greatly appreciate any and all help I can get.

a. Move the constant term to to the right side of the equation
b. Multiply each term in the equation by four times the coefficient of the x squared term.
c. Square the coefficient of the original x term and add it to both sides of the equation.
d. Take the square root of both sides.
e. Set the left side of the equation equal to the positive square root of the equation and solve for x
f. Set teh left side equal to the negative of the square root and solve for x.

this is the example they gave me.

solve x^+3x-10=0
x^+3x=10
4x^+12x=40
4x^+12x+9=40+9
4x^+12x+9=49
2x+3=+-7

2x+3=7 and 2x+3=-7
2x=4 2x=-10
x=2 x=-5

This seemed simple enough, but when I started trying it on my own with the problems they gave me, I am having a very difficult time. Below is listed one of the problems given. Can someone help me solve this, and give a little more detail when it gets to taking the square of both sides. I would greatly appreciat it. Thank you soooo much.

x^-2x-13=0

 

[soroban]

Re: quadratic equations

Hello, bbayes81!

\(\displaystyle x^2 - 2x - 13 \;=\;0\)

a. Move the constant term to to the right side of the equation
. . \(\displaystyle x^2 - 2x \:=\:13\)

b. Multiply each term in the equation by four times the coefficient of the x squared term.
. . \(\displaystyle 4x^2 - 8x \:=\:52\)

c. Square the coefficient of the original x term and add it to both sides of the equation.
. . \(\displaystyle 4x^2 - 8x + 4 \:=\:52 + 4\quad\Rightarrow\quad (2x-2)^2 \;=\;56\)

d. Take the square root of both sides.
. . \(\displaystyle 2x - 2 \;=\;\pm\sqrt{56} \;=\;\pm2\sqrt{14}\)

e. Set the left side of the equation equal to the positive square root and solve for x
. . \(\displaystyle 2x-2 \:=\:2\sqrt{14} \quad\Rightarrow\quad 2x \:=\:2 + 2\sqrt{14} \quad\Rightarrow\quad x \;=\;\frac{2 + 2\sqrt{14}}{2} \;=\;\boxed{1 + \sqrt{14}}\)

f. Set the left side of the equation equal to the negative square root and solve for x.
. . \(\displaystyle 2x-2 \:=\:-2\sqrt{14} \quad\Rightarrow\quad 2x \:=\:2 - 2\sqrt{14} \quad\Rightarrow\quad x \:=\:\frac{2-2\sqrt{14}}{2} \:=\:\boxed{1 - \sqrt{14}}\)

 

[stapel]

x^-2x-13=0

Note: While the carat character, "^", is the standard indicator for typed exponents, this character should be followed by what the power actually is. In this case, the equation "x[sup:r4asz7ga]2[/sup:r4asz7ga] - 2x - 13 = 0" would be written as "x^2 - 2x - 13 = 0".

when I started trying it on my own with the problems they gave me, I am having a very difficult time.

If you are still having trouble, please reply showing what you did and where you are having "a very difficult time". (We can't solve problems we can't see! :wink: )

Below is listed one of the problems given. Can someone...give a little more detail when it gets to taking the square of both sides.

Actually, you're not supposed to square both sides; you're supposed to take the square root of both sides. :shock:

They were supposed to have taught a section or two on solving radical equations first, so you'd be prepared for this method. Since you're not familiar with it, you might want to fill in what the book and lectures omitted by studying some online lessons. :idea:

. . . . .Google results for "solving radical equations"

. . . . .Google results for "solving quadratic equation square root"

Hope that helps a bit!

Eliz.

 

[Deleted member 4993]

I have not encountered that method before - and I have encountered many quadratic equations.

 

[stapel]

I have not encountered that method before...

You never had to complete the square to solve a quadratic? Nice! :wink:

Eliz.

 

[Deleted member 4993]

Re:

I have not encountered that method before...

You never had to complete the square to solve a quadratic? Nice! :wink:

Eliz.

I did - but I did the following way:

\(\displaystyle Ax^2 + Bx + C = 0\)

\(\displaystyle A(x^2 + \frac{B}{A}\cdot x + \frac{C}{A}) = 0\)

\(\displaystyle A(x^2 + 2\cdot\frac{B}{2A}\cdot x + [\frac{B}{2A}]^2 -[\frac{B^2}{4A^2} - \frac{C}{A}]) = 0\)

and so on...

I understand that the step of multiplying by 4 - takes away the reason for dividing by 4A^2 - but will only work for finding roots. If they need to deal with parabola (y=f(x)) - then they will have to learn whole another set of reasoning (not too different but similar enough to be confusing and making mistakes). I think we should call spade a spade - instead of trying to disguise it as "post-hole-digger".

 

[bbayes81]

Here is another example and I am still having trouble

4x^2-4x+3=0

4x^2-4x=-3

64x^2-64x=-48

64x^2-64x+16=-48+16

64x^2-64x+16=-32

this is where I am getting stuck. I need to take the square root of each side, but I am not sure how to do it.

Can you show me in detail what you do beyond this point??

 

[stapel]

64x^2-64x+16=-32

this is where I am getting stuck. I need to take the square root of each side, but I am not sure how to do it.

Convert the left-hand side from perfect-square form to squared-binomial form (like was covered back in "factoring", when you would be asked "is the following trinomial a perfect square?"); it might help, of course, to factor the common factor out first, giving you 16(4x[sup:1l5usy7r]2[/sup:1l5usy7r] - 4x + 1) = -32. :wink:

Eliz.