Verified answer

I assume that the x is squared

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(think of it as a tiny arrow pointing up, showing that the following number should be seen as if it were raised above the line)

2(x^2 + 5) = 60

Can it be x = 5 ?

we test:

2(5^2 + 5) = 2(25 + 5) = 2(30) = 60

it works, therefore x = 5 is a solution.

Can it be x = -5?

we test again:

2[(-5)^2 + 5) = 2(25 + 5) = 2(30) = 60

it works, therefore x = -5 is also a solution.

The way you did it, there could be some confusion between step 4 and step 5 (because you do not explain how you go from 4 to 5).

If you simply "take the square root" on both sides, someone could argue that the "principal" square root of a number is the positive value (+5 only, not -5). On the other hand, if you state:

(after step 4), we look for all values that give 25 when they are squared, then -5 is allowed.

The way the conjecture is stated at the beginning, the easiest way is to check both values to find out that, yes, they are both valid solutions.

2(x^2 + 5) = 60

2x^2 + 10 = 60

2x^2 = 50

x^2 = 25

x = 5.

problem: 2(x^2+5)=60 step a million: distribute 2x^2+10=60 step 2: subtract 10 2x^2=50 step 3: divide via 2 x^2=25 step 4: take the sq. root x=5 or x= -5 X might properly be the two 5 or -5 based on your area. area is the values of x you need to use on your problem. in the experience that your area is infinity to infinity then your answer is the two. in case you area is from 0 to infinity then your answer is 5. in the experience that your area is damaging infinity to 0 then your answer is -5. i wish this facilitates!

Your working is right. The answer is plus or minus 5 :)

x² + 5 = 30

x² = 25

x = ± 5________As you suggest.

yes