Inter. Algebra: Complete square for 2x^2 + 3x = 2 + 5x

[Helen]

Complete the square. Then (i) give the exact solutions and (ii) give solution rounded to nearest thousandth.
This is the problem: 2x ^ (2) + 3x = 2 + 5x
My answer: (i) {1 (+minus sign under the +) sqrt(5)} (ii) {3.236, - 1.236}
Is my answer correct?

 

[o_O]

Re: Intermediate Algebra

Think you forgot to divide your answer by 2:

\(\displaystyle x = \frac{1 \pm \sqrt{5}}{2}\)

And again, you can always check your answer yourself by plugging it back into the original equation.

Look at your solution x = 3.236.
\(\displaystyle \mbox{Left hand side: } 2x^{2} + 3x = 2(3.236)^{2} + 3(3.236) = 2(10.471696) + 3(3.236) = 20.943392 + 9.708 = 30.651392\)
\(\displaystyle \mbox{Right hand side: } 2 + 5x = 2 + 5(3.236) = 2 + 16.18 = 18.18\)

We can see that both sides are not equal so the solution isn't correct. If it was correct, then both sides would have the same numeric value.

 

[Helen]

Re: Intermediate Algebra

,
How about this? (i) {1 (+ with - sign under +) sqrt 5} (ii) {1.618, -0.618}
over 2

 

[o_O]

Re: Intermediate Algebra

Why don't you plug it in to your original equation to see if they're approximately the same like I showed you ?

 

[Mrspi]

Re: Intermediate Algebra

Complete the square. Then (i) give the exact solutions and (ii) give solution rounded to nearest thousandth.
This is the problem: 2x ^ (2) + 3x = 2 + 5x
My answer: (i) {1 (+minus sign under the +) sqrt(5)} (ii) {3.236, - 1.236}
Is my answer correct?

Your original equation is

2x[sup:3atnhu9x]2[/sup:3atnhu9x] + 3x = 2 + 5x

Get one side to equal 0. I'll add -2 - 5x to both sides of the equation:

2x[sup:3atnhu9x]2[/sup:3atnhu9x] + 3x - 2 - 5x = 2 + 5x - 2 - 5x

2x[sup:3atnhu9x]2[/sup:3atnhu9x] -2x - 2 = 0

Divide both sides of the equation by 2:

x[sup:3atnhu9x]2[/sup:3atnhu9x] - x - 1 = 0

There's your equation. It's a quadratic equation so I'd suggest you use the quadratic formula to solve it (the left side does not factor over the integers.)

 

[Helen]

Re: Intermediate Algebra

,
I followed your advice and by plugging into the original equation, got this answer.
(i) {-1 ) (+ with the - sign under the +) sqrt 5} (ii) {1.236, -3 . 236}
They matched, so I believe it is right. Thank you for showing me how to plug in.
Could you look it over just to make sure and I will check on it as soon as I return.Thanks again!
Helen

 

[Helen]

Re: Intermediate Algebra

mrspi,
Thank you for your help. I printed out both yours and 's paperwork so that I can refer to it later. I do appreciate it. Helen

 

[o_O]

Re: Intermediate Algebra

Yep they should be right. Just plug it in back to the original equation and just see if the equation is actually an equality!

 

[Denis]

Re: Intermediate Algebra

,
I followed your advice and by plugging into the original equation, got this answer.
(i) {-1 ) (+ with the - sign under the +) sqrt 5} (ii) {1.236, -3 . 236}
They matched, so I believe it is right. Thank you for showing me how to plug in.
Could you look it over just to make sure and I will check on it as soon as I return.Thanks again!
Helen

NO. Your -1 should be +1; plus you need to divide by 2 !

x = [1 +- sqrt(5)] / 2 : that'll give you x = 1.618 or x = -.618

AND start using +- for: (+ with the - sign under the +)

 

[o_O]

Re: Intermediate Algebra

Oh whoops. I assumed you got it right seeing how you posted the correct solution in the 3rd post.

 

[Helen]

Re: Intermediate Algebra

Denis, Thank you for the correction. Helen