# Solving the equation $x + \sqrt{2x+1} = 7$

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.

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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

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).

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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
@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
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 $x\le 7$.)

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

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

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

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

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

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

$x=12,\ x=4$

$S={4}$

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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 $x = 12$.

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