How do you solve #4x^2=28#?

Question

Kennedy Adams · Accepted Answer

#x^2=7# or #x=sqrt7# Divide both sides by 4 (first step): #x^2=28/4# #x^2=7# Take square root (of both sides): #x=sqrt7# This is your answer: #x=sqrt7#

Abigail Allen · Answer

#x=+-sqrt7# #color(blue)"Isolate " x^2" by dividing both sides by 4"# #(cancel(4) x^2)/cancel(4)=28/4# #rArrx^2=7# #color(blue)"take the square root of both sides"# remembering that the square root of a number can have a positive/negative value. #sqrt(x^2)=color(red)(+-)sqrt7# #rArrx=color(red)(+-)sqrt7#

HIX Tutor · Answer

undefined Great question! Let’s solve $4x^2 = 28$ step-by-step.

1. **What do you notice about both sides of the equation?** 
   - There’s a number on each side.

2. **Can we make it simpler?** 
   - Yes! Let’s divide both sides by 4 to get $x^2$ alone.

$$
   x^2 = \frac{28}{4}
   $$

3. **What does $\frac{28}{4}$ equal?**
   - Yes, it equals 7. So now we have:

$$
   x^2 = 7
   $$

4. **What do we want to find now?**
   - We want to find $x$.

5. **What do we do to get $x$ by itself?** 
   - We can take the square root of both sides.

$$
   x = \sqrt{7}
   $$

6. **Does $x$ have another value?**
   - Yes! We also have a negative value, so:

$$
   x = \sqrt{7} \quad \text{or} \quad x = -\sqrt{7}
   $$

7. **What are the final answers?**
   - The answers are $x = \sqrt{7}$ and $x = -\sqrt{7}$.

Does that help you see how to solve it?

Ethan Adkins · Answer

undefined To solve $4x^2 = 28$:

1. Divide both sides of the equation by 4.
2. Simplify the equation.
3. Take the square root of both sides of the equation.
4. Solve for $x$.

$$4x^2 = 28$$

$$x^2 = \frac{28}{4}$$

$$x^2 = 7$$

$$x = \pm \sqrt{7}$$

So, the solutions are $x = \sqrt{7}$ and $x = -\sqrt{7}$.