Hi Tracey,
You could expand the left side of this equation, collect the terms to get a quadratic equation in x and either factor it or use the general quadratic to solve for x. As with every problem however you should look at the problem closely and ask if it makes sense. In the problem you have a square on one side of the equal sign and a negative number on the other side. Can a square be negative?
Penny
Hi Tracey.
The product of any negative number times itself is positive, and of course so is a positive number times itself. What this means of course is that all real numbers squared are positive. So your question (4x5)^{2 } = 32 doesn't have a "real" solution.
In such a case, we have "imaginary numbers". That's where you see the letter i attached to real numbers, as in 4 + 5i which means literally 4 plus 5 times the square root of 1.
In your question, let's multiply it out to a quadratic equal to zero:
(4x5)^{2} + 32 = 0
16x^{2} 40x +25 + 32 = 0
16x^{2}  40x + 57 = 0
Now if you use the quadratic equation you get a=16, b=40 and c=57,
so
x = [ b +/ sqrt( b^{2 }  4ac) ] / 2a
x = [ (40) +/ sqrt[ (40)^{2}4(16)(57) ] ] / (2(16))
x = [ 40 +/ sqrt(2048) ] / 32
x = [ 40 +/ 32sqrt(2) ] / 32
x = 5/4 +/ sqrt(2)sqrt(1)
x = 1.25 +/ sqrt(2)i
So there are two imaginary solutions: 1.25 + sqrt(2)i and 1.25  sqrt(2)i.
Whenever I do this kind of algebra, I like to check my answer by putting it in for x in the original equation:
(4x5)^{2 }
=(4 (1.25 +sqrt(2)i )  5)^{2}
= (5 + 4sqrt(2)i  5)^{2}
= (4sqrt(2)i)^{2}
= (4^{2 })(2)i^{2}
= 32i^{2}
and since i is the square root of 1, then i^{2 } is 1, so the answer is 32, which is what we were hoping for. You can do the same with the second answer (1.25  sqrt(2)i) and you'll also get 32.
Hope this helps,
Stephen La Rocque.
