Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I can't solve

$x + \sqrt{2x+1} = 7$.

Well, I know the answer is 4, but that is from just reasoning it out. I can't algebraically solve it.

Thus, a step by step is what I really need.

Thanks in advance!!

share|improve this question
3  
Start by isolating the square root on one side $\sqrt{2x+1}=7-x$. Then square both sides of the equation. You'll get a quadratic equation in $x$. –  Henning Makholm Sep 15 '11 at 16:23

4 Answers 4

up vote 5 down vote accepted

There are several ways of doing this. Here's one: set $u=\sqrt{2x+1}$. Then $u^2 = 2x+1$, so $2x = u^2-1$, $x=\frac{1}{2}(u^2-1)$. Substituting into the original equation, you have $$\frac{1}{2}(u^2-1) + u = 7$$ which is a quadratic equation in $u$: $$u^2 +2u -15 = 0.$$ We can solve this equation for $u$. Then solve $u=\sqrt{2x+1}$ for $x$, remembering that $u$ must be nonnegative, and $x$ must be greater than or equal to $-\frac{1}{2}$. Plug into the original equation to verify they give you correct answers.

Another method (essentially the same, but without introducing new variables): Rewrite as $$\sqrt{2x+1} = 7-x.$$ Square both sides, solve for $x$, then plug back into the original equation to verify (squaring both sides may introduce "extraneous solutions", so you need to check the answers you got actually solve the original).

share|improve this answer
    
Thanks!! I learned a few things here. Quick additional question, When I square an equation, do I have to square both sides, as opposed to squaring every term individually? –  Kuba Sep 15 '11 at 16:38
2  
@Kuba: You do not square each term individually; you have to square the entire side, and do it on both sides. For instance, $3+3=6$, but if you square "each term individually", you would get $9+9=36$, which is certainly wrong! –  Arturo Magidin Sep 15 '11 at 16:40
1  
wow I actually just learned a few things that I should've known by now! You should be a teacher...if you're not already. Thanks again!!! –  Kuba Sep 15 '11 at 16:42
    
+1 for the very useful tip "squaring both sides may introduce "extraneous solutions", so you need to check the answers you got actually solve the original" –  Quixotic Sep 28 '11 at 0:34

$7-x= \sqrt{2x+1}$

$ (7-x)^2 = 2x+1$

$ x^2-14x+49 = 2x+1 $

$x^2-16x+48=0$

$(x-8)^2-64+48=0$

$(x-8)^2=16$

$x-8=\pm 4$

$x=12$ or $x=4$

But $x=12$ does not work in the original equation. So the answer is $x=4$. (Or the original equation requires $7-x\ge 0$ and so $x\le 7$.)

share|improve this answer
    
thanks! It doesn't get any more step-by-step than this! –  Kuba Sep 15 '11 at 16:38

$x+\sqrt{2x+1}=7$

$(x-7)^2=2x+1$

$x^2-14x+49-2x-1=0$

$x^2-16x+48=0$

$(x-12)(x-4)=0$

$x=12,\ x=4$

$S={4}$

share|improve this answer
    
we lost the {4} in the edit... –  The Chaz 2.0 Sep 27 '11 at 21:51

If $x + \sqrt{2x+1} = 7$, multiply both sides by $x - \sqrt{2x+1}$ to get $x^2 - (2x+1) = 7(x - \sqrt{2x+1})$, or $$\sqrt{2x+1} = x - (x^2 - (2x+1))/7 = (-x^2 + 9x+1)/7.$$ Substituting in the original equation, $7 = x + (-x^2 + 9x+1)/7 = (-x^2 + 16x+1)/7$ or $x^2 - 16x + 48 = 0$. Solving this, we get $x = 4$ and $x = 12$. $x = 4$ satisfies the original equation with a positive square root, and $x = 12$ satisfies it with a negative square root.

You also have to make sure that the multiplying by $x - \sqrt{2x+1}$ does introduce any extraneous root(s). Since doing this throws away the sign of the square root, this introduces the value $x = 12$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.