Question

solve the radical equation sqrt -7x+4 = sqrt 1-x +3

Answer 1

To solve the equation

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

we first isolate one of the square root terms. We'll start by isolating the square root on the left:

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

Next, we can square both sides to eliminate the square roots:

\[
\left(\sqrt{-7x + 4} - 3\right)^2 = (1 - x).
\]

Expanding the left side gives us:

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

This simplifies to:

\[
-7x + 13 - 6\sqrt{-7x + 4} = 1 - x.
\]

Rearranging terms leads to:

\[
-7x + 13 - 1 + x = 6\sqrt{-7x + 4},
\]

which simplifies to:

\[
-6x + 12 = 6\sqrt{-7x + 4}.
\]

Dividing everything by 6 gives us:

\[
-x + 2 = \sqrt{-7x + 4}.
\]

Now, we square both sides again:

\[
(-x + 2)^2 = -7x + 4.
\]

Expanding the left side results in:

\[
x^2 - 4x + 4 = -7x + 4.
\]

We can simplify this by subtracting 4 from both sides:

\[
x^2 - 4x = -7x.
\]

Rearranging gives us:

\[
x^2 + 3x = 0.
\]

Factoring out \(x\):

\[
x(x + 3) = 0.
\]

This gives us the solutions:

\[
x = 0 \quad \text{or} \quad x = -3.
\]

Next, we must check both solutions in the original equation for any extraneous solutions.

1. **Checking \(x = 0\)**:

\[
\sqrt{-7(0) + 4} = \sqrt{1 - 0} + 3,
\]
\[
\sqrt{4} = \sqrt{1} + 3,
\]
\[
2 = 1 + 3,
\]
\[
2 = 4 \quad \text{(false)}.
\]

So, \(x = 0\) is not a valid solution.

2. **Checking \(x = -3\)**:

\[
\sqrt{-7(-3) + 4} = \sqrt{1 - (-3)} + 3,
\]
\[
\sqrt{21 + 4} = \sqrt{4} + 3,
\]
\[
\sqrt{25} = 2 + 3,
\]
\[
5 = 5 \quad \text{(true)}.
\]

So, \(x = -3\) is a valid solution.

Thus, the solution to the equation is:

\[
\boxed{-3}.
\]