Solve the radical equation −7x+4−−−−−−−√=1−x−−−−√+3 .(1 point) x =

Answer 1

To solve the equation $-\sqrt{7x+4}=\sqrt{1-x}+3$, we start by isolating the square roots on one side:

1. Move all non-square root terms to one side: $$-\sqrt{7x+4}-3=\sqrt{1-x}$$

2. Next, we square both sides to eliminate the square roots. Remember: when squaring both sides, we need to be careful about extraneous solutions: $$(-\sqrt{7x+4}-3{)}^2=(\sqrt{1-x}{)}^2$$

3. Expanding both sides gives us: $$(7x+4)+6\sqrt{7x+4}+9=1-x$$ $$7x+13+6\sqrt{7x+4}=1-x$$

4. Rearranging it gives: $$6\sqrt{7x+4}=1-x-7x-13$$ $$6\sqrt{7x+4}=-8x-12$$

5. Dividing both sides by 6: $$\sqrt{7x+4}=\frac{-8x-12}{6}$$ Simplifying the right side: $$\sqrt{7x+4}=-\frac{4x+6}{3}$$

6. Now, we square again to eliminate the square root: $$7x+4={(-\frac{4x+6}{3})}^2$$ $$7x+4=\frac{(4x+6{)}^2}{9}$$

7. Multiplying through by 9 to clear the fraction: $$9(7x+4)=(4x+6{)}^2$$ $$63x+36=16x^2+48x+36$$

8. Simplifying gives: $$0=16x^2-15x$$ Factoring out $x$: $$0=x(16x-15)$$

9. Setting each factor to zero gives: $$x=0\text{or}16x-15=0⟹x=\frac{15}{16}$$

Now we must check for extraneous solutions by substituting back into the original equation.

For $x=0$: $$-\sqrt{7(0)+4}=\sqrt{1-0}+3$$ $$-2=1+3⟹-2\ne 4(\text{not a solution})$$

For $x=\frac{15}{16}$: $$-\sqrt{7\left(\frac{15}{16}\right)+4}=\sqrt{1-\frac{15}{16}}+3$$ Calculating: $$7\left(\frac{15}{16}\right)+4=\frac{105}{16}+\frac{64}{16}=\frac{169}{16}⟹-\frac{13}{4}$$ $$\frac{1}{4}+3=\frac{1}{4}+\frac{12}{4}=\frac{13}{4}$$ So, $$-\frac{13}{4}\ne \frac{13}{4}(\text{not a solution})$$

Since neither solution is valid in the original equation, the equation has no solution.

Thus the final answer is: $$\boxed{{\displaystyle \text{No solution}}}$$