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

Question

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?

Answer 1

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.

Answer 2

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

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

Answer 3

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 $\sqrt{3}>1$.

Answer 4

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.

Answer 5

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. $