How do you solve #10b ^ { 2} - 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 = ± sqrt(105)/5 i#.

Explanation:

We have:
#10b^2 - 7 = -49#

We could add #7# to both sides:

#10b^2 - 7 + 7 = -49 + 7#

#10b^2= -42#

And divide by #10#:

#(10b^2)/10 = (-42)/10#

#b^2 = -21/5#

Then take the square root of #b^2#, but be careful! If #3^2 = 9# and #(-3)^2 = 9# then #sqrt(9) = ±3#.

#sqrt(b^2) = ±sqrt(-21/5)#

#b = ±sqrt(-21/5)#

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 = ± sqrt(21/5) i#

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

#b = ± (sqrt(21))/sqrt(5) i#

Rationalize the denominator:

#b = ± (sqrt(21) * sqrt(5))/(sqrt(5) * sqrt(5)) i#

#b = ± sqrt(105)/5 i#

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

#b_1 = sqrt(105)/5 i#

#b_2 = -sqrt(105)/5 i#

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

Related questions
Impact of this question
916 views around the world
You can reuse this answer
Creative Commons License