# Solving $\sqrt{x - 4} + \sqrt{x - 7} = 1$.

I have the equation $\sqrt{x - 4} + \sqrt{x - 7} = 1$.

I tried to square both sides, but then I got a more difficult equation: $$2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.$$ Can someone tell me what I should do next?

• Get everything but the root to the other side and square again. – Gregory Grant May 13 '15 at 17:57
• Putting 8 in the first equation, this leads you to 3=1, check the answers before accepting one. – Renato Faraone May 13 '15 at 18:34
• Note that squaring an equation like this can give spurious solutions, because squaring loses information about signs. So the putative solution identified below i.e. $x=8$ which doesn't work, is a solution of the related equation $\sqrt{x-4}-\sqrt {x-7}=1$ – Mark Bennet May 13 '15 at 18:34

squaring gives $$x-4+x-7+2\sqrt{x-4}\sqrt{x-7}=1$$ $$\sqrt{x-4}\sqrt{x-7}=12-2x$$ $$(x-4)(x-7)=36+x^2-12x$$ $$x=8$$ but $8$ fulfills not our equation.

• Of course, you have to check if this is a solution, which it isn't. – ET93 May 13 '15 at 18:02
• i have checked it – Dr. Sonnhard Graubner May 13 '15 at 18:04
• Try to put $8$ in the first equation, this leads you to $3=1$ which is clearly false! – Renato Faraone May 13 '15 at 18:31

There is no real solution since you need $x \geq 7$ for sure. The left hand side is an increasing function for $x \geq 7$ and so it's minimum is $\sqrt{3}>1$.

• +1: A solution of supreme elegance without any hard calculations. – Berrick Caleb Fillmore May 13 '15 at 18:04

Suppose that $x$ is a solution. Flip things over. We get $$\frac{1}{\sqrt{x-4}+\sqrt{x-7}}=1.$$ Rationalizing the denominator, and minor manipulation gives $$\sqrt{x-4}-\sqrt{x-7}=3.$$ From this, addition gives $2\sqrt{x-4}=4$. The only possibility is $x=8$, which is not a solution.

$$\sqrt{x-4}+\sqrt{x-7}=1<=>$$ $$-11+2\sqrt{(x-7)(x-4)}=12-2x<=>$$ $$4(x-7)(x-4)=(12-2x)^2<=>$$ $$4x^2-44x+112=4x^2-48x+144<=>$$ $$4x-32=0<=>$$ $$4(x-8)=0<=>$$ $$x-8=0<=>$$ $$x=8$$

BUT THIS SOLUTION IS INCORRECT SO THERE ARE NO SOLUTIONS!!!!!

• If this solution is incorrect, you shouldn't use equivalence arrows in each step! – Chris May 13 '15 at 18:35
• When you square both sides, you use implication arrows, not iff, unless you add some additional inequality constraints for $x$. Squaring is not an invertible operation unless with some additional inequality constraints. And indeed $8$ is not a solution, so you can't have iff arrows everywhere. – user26486 May 14 '15 at 2:53
• My math professor on the university told us that we have to use them to give the relationship between the equations, so here in Holland we use them as I did – Jan May 14 '15 at 7:24
• @JanEerland Algebraic manipulations often result in equivalent expressions. But not in the case of squaring both sides. E.g., we cannot write $x=2\!\iff\! x^2=4$, because $(-2)^2=4$ while $-2\neq 2$. – user26486 May 14 '15 at 10:42

We cannot assume that a root/solution always exists, especially when sqrt is present. At first test with a few values of $x$ to ascertain whether it can be satisfied approximately. The squaring sometimes introduces possibility of the second sign before the radical.

The equation as you gave ( positive value for both sqrts) has no real solution as x must be > 7.

However, $$-\sqrt{x-7} + \sqrt{x-4} = 1$$ has the solution $x = 8.$