Find the real numbers a,b,c if a^2+b^2+c^2-4a-6b+10c+38=0?

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shrinath's profile pic

shrinath | Student, Undergraduate | (Level 1) Honors

Posted on

a^2+b^2+c^2-4a-6b+10c+38=0

=>(a-4)a + (b-6)b+c(c+10)+38 =0

(a-2)^2 + (b-3)^2 + (c+5)^2=0

a^2 +b^2+c^2+10c+38 = 4a + 6b

Let value of a and b be

a=2 and b=3  c=?

c = sqrt (-a^2 +4a -b ^2+6b -13)

=> c =-5

Substituting so on we have

a= 2, b=3, c=-5

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-group the terms to emphasize the fact that we'll have to complete a number of squares:

(a^2 - 4a) + (b^2 - 6b) + (c^2 + 10c) + 38 = 0

We'll have to complete the squares inside the brackets:

(a^2 - 4a + 4) + (b^2 - 6b + 9) + (c^2 + 10c + 25) - 4 - 9 - 25 + 38 = 0

(a^2 - 4a + 4) + (b^2 - 6b + 9) + (c^2 + 10c + 25) - 38 + 38 = 0

We'll eliminate like terms and we'll recognize the perfect squares:

(a-2)^2 + (b-3)^2 + (c+5)^2 = 0

The sum of the squares cannot be zero, unless the value of each square is zero.

a - 2 = 0 <=> a = 2

b - 3 = 0 <=> b = 3

c + 5 = 0 <=> c = -5

The requested values of the numbers a,b,c, for the given identity to hold, are: a = 2 , b = 3 , c = -5.

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