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If a+b+c=9 and ab+bc+ca=23, then a2+b2+c2=?
127
35
25
58
35
Use the identity (a+b+c)2=∑a2+2∑ab. Simply calculate 92−2(23)=81−46=35.
a + b + c = 9, ab + bc + ca = 23
(a+b+c)2=a2+b2+c2+2(ab+bc+ca)
Use the identity (a+b+c)2=∑a2+2∑ab. Simply calculate 92−2(23)=81−46=35.
A common mistake is to forget the factor of 2 when substituting 2(ab+bc+ca), which leads to 81−23=58 (a distractor option).
State the algebraic identity
To find a2+b2+c2, we use the standard expansion for the square of a trinomial.
(a+b+c)2=a2+b2+c2+2(ab+bc+ca)
Substitute the given values
Given a+b+c=9 and ab+bc+ca=23. Substitute these values into the identity: (9)2=a2+b2+c2+2(23).
81=a2+b2+c2+46
Solve for the unknown sum of squares
Isolate a2+b2+c2 by subtracting 46 from both sides of the equation.
a2+b2+c2=81−46=35
B is correct because the calculation 81−46 results in 35.
This identity is fundamental for problems involving systems of symmetric equations and is frequently tested in exams alongside Vieta's formulas for polynomials.
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