Solve the radical equation −7x+4−−−−−−−√=1−x−−−−√+3

Question

.(1 point)
x=

Answer 1

To solve the equation

$$-\sqrt{-7x+4}=\sqrt{1-x}+3,$$

we first isolate the square root on one side by adding $\sqrt{-7x+4}$ and $\sqrt{1-x}$ to both sides:

$$0=\sqrt{1-x}+3+\sqrt{-7x+4}.$$

Now, let's rearrange it to:

$$\sqrt{-7x+4}=-\sqrt{1-x}-3.$$

Since the square root on the left side is non-negative, the right side must also be non-negative. Thus:

$$-\sqrt{1-x}-3\ge 0,$$

which implies

$$-\sqrt{1-x}\ge 3\Rightarrow \sqrt{1-x}\le -3,$$

which is impossible, because square roots are always non-negative. This indicates that either:

1. There are no solutions, or
2. We made a mistake in how we isolated the terms.

Let's square both sides at some earlier point and see if we can solve it that way.

Starting from the original equation:

$$-\sqrt{-7x+4}=\sqrt{1-x}+3,$$

we can square both sides. Square both sides gives:

$$(-\sqrt{-7x+4}{)}^2=(\sqrt{1-x}+3{)}^2.$$

This simplifies to:

$$-7x+4=(1-x)+6\sqrt{1-x}+9.$$

Updating gives:

$$-7x+4=10-x+6\sqrt{1-x}.$$

Rearranging further,

$$-7x+x+4-10=6\sqrt{1-x}⟹-6x-6=6\sqrt{1-x}.$$

Dividing by 6 yields:

$$-x-1=\sqrt{1-x}.$$

Again, we square both sides:

$$(-x-1{)}^2=1-x.$$

Expanding gives:

$$x^2+2x+1=1-x.$$

Bringing everything to one side gives:

$$x^2+3x=0.$$

Factoring out $x$:

$$x(x+3)=0.$$

Thus, we have two potential solutions:

$$x=0\text{or}x=-3.$$

Now, we need to check whether these solutions satisfy the original equation:

For $x=0$

$$-\sqrt{-7(0)+4}=\sqrt{1-0}+3⟹-\sqrt{4}=1+3⟹-2=4\text{(false)}.$$

For $x=-3$

$$-\sqrt{-7(-3)+4}=\sqrt{1-(-3)}+3⟹-\sqrt{21+4}=\sqrt{4}+3⟹-\sqrt{25}=2+3⟹-5=5\text{(false)}.$$

Since neither solution satisfies the original equation, the equation $-\sqrt{-7x+4}=\sqrt{1-x}+3$ has no solutions in real numbers.

Thus, the answer to the original equation is:

$$\boxed{{\displaystyle \text{No solutions}}}.$$