Question #b0e7d

2 Answers
Zack M.
Aug 22, 2017

#a-b-c = 10#

Explanation:

We are given two equations and asked to solve for a third.

#a^2 + b^2 + c^2 = 30#
#bc - ca - ab = 35#
#a - b - c = ?#

In order to do so, we need to find some sort of relationship between all three equations. Since there are squares in the first equation, lets try squaring the unsolved equation.

#a - b - c = sqrt((a - b - c)^2)#

If we expand the part underneath the radical, we get;

#=sqrt((a^2 - ab - ac - ab + b^2 + bc - ac + bc + c^2))#

Now all of the terms underneath the radical are the same as the terms in the given equations. After rearranging, we have;

#=sqrt((a^2 + b^2 + c^2 + 2bc - 2ac - 2ab))#

Adding parenthesis, we can see the first two equations under the radical.

#=sqrt(((a^2 + b^2 + c^2) + 2(bc - ac - ab)))#

We can replace the terms within parenthesis with their solved values from above.

#=sqrt((30 + 2(35)))#

Finally, solve the expression.

# = sqrt(100)#

# = 10#

Cesareo R.
Aug 22, 2017

#a-b-c = pm 10#

Explanation:

#(a - b - c)^2 = (a^2 + b^2 + c^2)+2(bc-ac-ab)# so

#(a-b-c)^2=30+2 xx 35=100 = 10^2# then

#a-b-c = pm 10#

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