How do you solve 10b2−7=−49?

1 Answer
Bisma J.
Dec 14, 2017

Solve algebraically to find that there are no real solutions, but there are complex solutions of the form b=±√1055i.

Explanation:

We have:
10b2−7=−49

We could add 7 to both sides:

10b2−7+7=−49+7

10b2=−42

And divide by 10:

10b210=−4210

b2=−215

Then take the square root of b2, but be careful! If 32=9 and (−3)2=9 then √9=±3.

√b2=±√−215

b=±√−215

Uhh, I'm afraid there is no real solution. We could, however, continue to find complex solutions, by first turning the negative symbol in the root into i, in a sense:

b=±√215i

Then solve for the root normally. Let's split it into the numerator and the denominator:

b=±√21√5i

Rationalize the denominator:

b=±√21⋅√5√5⋅√5i

b=±√1055i

Now, the plus-or-minus sign there indicates that it could be positive or negative, so we have two solutions:

b1=√1055i

b2=−√1055i

That's as far as we can go without evaluating the square root.

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