Question

Find the solution set.
√(x+24)-√(x-16)=4

Answer 1

Let's solve your equation step-by-step.
x+24−x−16=4
Step 1: Square both sides.
(x+24)2−2x+24x−16+(x−16)2=16
x+24−2(x+24)(x−16)+x−16=16
−2x2+8x−384+2x+8=16
Step 2: Add -8 to both sides.
−2x2+8x−384+2x+8+−8=16+−8
−2x2+8x−384+2x=8
Step 3: Add -2x to both sides.
−2x2+8x−384+2x+−2x=8+−2x
−2x2+8x−384=−2x+8
Step 4: Divide both sides by -2.
−2x2+8x−384−2=−2x+8−2
x2+8x−384=x−4
Step 5: Solve Square Root.
x2+8x−384=x−4
x2+8x−384=(x−4)2(Square both sides)
x2+8x−384=x2−8x+16
x2+8x−384−x2=x2−8x+16−x2(Subtract x^2 from both sides)
8x−384=−8x+16
8x−384+8x=−8x+16+8x(Add 8x to both sides)
16x−384=16
16x−384+384=16+384(Add 384 to both sides)
16x=400
16x16=40016(Divide both sides by 16)
x=25
Check answers. (Plug them in to make sure they work.)
x=25(Works in original equation)
Answer:
x=25

Answer 2

love and appreciate the work but (x+4) and (x-16) are both in radicals. if you square everything you'd be down to x+24-x-16=16 or 8=16 because the x's cancel out. that's why I'm stuck on this problem

Answer 3

@touseii45 Umm, if you are squaring a sum of 2 different variables, then (x+y)^2 is NOT equal to x^2+y^2. You need to apply the formula x^2+y^2+2xy

Answer 4

i think he copied the entire solution from somewhere
but it is correct

Answer 5

@touseii45 if you do not believe what I wrote, then take any two numbers and you will see. for example 3^2+4^2 = 25, while (3+4)^2=49

Answer 6

i don't think i understand.

Answer 7

are you saying that √(x+24) being squared would be seen as x^2 + 24^2 and √(x-16) would be seen as x^2 -16^2 ?