Need help associated with the following equation

[gamaz321]

Hello,
I have an equation in a radical form. I could do a few steps here. However, it is becoming too complicated to manage after a few steps. I am wondering
if this has any better approach to deal with this particular problem.
Thank you.

 

[Dr.Peterson]

Hello,
I have an equation in a radical form. I could do a few steps here. However, it is becoming too complicated to manage after a few steps. I am wondering
if this has any better approach to deal with this particular problem.
Thank you.

That's an interesting trick you're using, but I see another.

​

To make the equation simpler, start with a substitution: let y = x^2 + x + 4. See what happens.

 

[BigBeachBanana]

Hello,
I have an equation in a radical form. I could do a few steps here. However, it is becoming too complicated to manage after a few steps. I am wondering
if this has any better approach to deal with this particular problem.
Thank you.

Square both sides, and simplify. It will give you a nice quartic that can be factored.
[math](\sqrt{a}+\sqrt{b})^2=(\sqrt{c})^2\\ a+b+2\sqrt{ab}=c [/math]
Edit: Making the substitution Dr P suggested before squaring both sides will simplify the problem into quadratic rather than quadric.

 

[gamaz321]

Thanks for the help to both of you. I tried applying both the suggestions. Unfortunately, I am still stuck. Here is my work in the attached document.

 

[Dr.Peterson]

Thanks for the help to both of you. I tried applying both the suggestions. Unfortunately, I am still stuck. Here is my work in the attached document.

Check your work solving the quadratic at the end. You could just factor it.

 

[gamaz321]

Thanks. The answer to the problem is 0 and -1. So putting 0 or -1 in the last equation where I quit does not seems to give us LHS = RHS. Maybe, I am missing some point?

 

[Dr.Peterson]

Thanks. The answer to the problem is 0 and -1. So putting 0 or -1 in the last equation where I quit does not seems to give us LHS = RHS. Maybe, I am missing some point?

The last equation is an equation in y, not x. Solve it for y, then find x.

 

[BigBeachBanana]

Thanks. The answer to the problem is 0 and -1. So putting 0 or -1 in the last equation where I quit does not seems to give us LHS = RHS. Maybe, I am missing some point?

Those are the correct answers for x. I'm not sure which equation you're referring to, but they work for the original equation.

Edit: Were those answers given to you or did you solve for them?

 

[gamaz321]

Is the value of y that I got in the screenshot in radical form correct? Here are a few more steps that I could do but running into a complicated situation to handle.

 

[BigBeachBanana]

Is the value of y that I got in the screenshot in radical form correct? Here are a few more steps that I could do but running into a complicated situation to handle.

I thought you understood what Dr.P said in #5. You didn't evaluate the quadratic formula correctly, so check it again. Alternatively, you can factor [imath]y^2+3y-4=(y\pm?)(y\pm?)=0[/imath], without using the quadratic formula.

 

[gamaz321]

Thanks, now it works. I was repeatedly making a silly mistake in calculating the (b^2 - 4ac) value of the quadratic equation of y. When I fixed it things are all good.
Best regards to you and Dr. P!

 

[BigBeachBanana]

It's interesting that the 2 complex solutions get washed out with the substitution.
[math] \sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\\ \text{Squaring both sides:}\\ (x^2+x+4)+(x^2+x+1)+2\sqrt{(x^2+x+4)(x^2+x+1)}=2x^2+2x+9\\ 2\sqrt{x^4 + 2 x^3 + 6 x^2 + 5 x + 4}=4\\ x^4 + 2 x^3 + 6 x^2 + 5 x =0\\ x (x + 1) (x^2 + x + 5) = 0\\ \implies x=0,-1,\frac{-1\pm\sqrt{19}i}{2} [/math]

 

[gamaz321]

It's my bad. I believe the question was to find the real values of x. Hence the imaginary values are not in the answers.

 

[BigBeachBanana]

It's my bad. I believe the question was to find the real values of x. Hence the imaginary values are not in the answers.

I inferred that much. I'm just including it for the sake of completeness.

Side note: Use the "reply" button in the bottom right corner so we know who you're replying to.

 

[gamaz321]

OK sure, thanks.

 

[Dr.Peterson]

It's interesting that the 2 complex solutions get washed out with the substitution.
[math] \sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\\ \text{Squaring both sides:}\\ (x^2+x+4)+(x^2+x+1)+2\sqrt{(x^2+x+4)(x^2+x+1)}=2x^2+2x+9\\ 2\sqrt{x^4 + 2 x^3 + 6 x^2 + 5 x + 4}=4\\ x^4 + 2 x^3 + 6 x^2 + 5 x =0\\ x (x + 1) (x^2 + x + 5) = 0\\ \implies x=0,-1,\frac{-1\pm\sqrt{19}i}{2} [/math]

No, the substitution method (as the OP did it) yields all four solutions; you get y=1 or -4, and the latter yields the two imaginary solutions, if you want them.

 

[BigBeachBanana]

No, the substitution method (as the OP did it) yields all four solutions; you get y=1 or -4, and the latter yields the two imaginary solutions, if you want them.

Oops. For some reason, I thought it yields the two 0 and -1 answers. That makes more sense since there are 3 quadratics.

 

[BigBeachBanana]

Oops. For some reason, I thought it yields the two 0 and -1 answers. That makes more sense since there are 3 quadratics.

Typo. I meant 2 quadratics.