Math Problem Statement

if a+b+c = 0 and a^2 + b^2 + c^2 = sqrt(74) what is a^4 + b^4 + c^4? Explain your working step by step

Solution

We are tasked to find a4+b4+c4a^4 + b^4 + c^4 given the conditions:

  1. a+b+c=0a + b + c = 0
  2. a2+b2+c2=74a^2 + b^2 + c^2 = \sqrt{74}

Step 1: Expand and simplify

The sum a4+b4+c4a^4 + b^4 + c^4 can be written using the following relation: a4+b4+c4=(a2+b2+c2)22(a2b2+b2c2+c2a2)a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2)

Thus, we need:

  1. a2+b2+c2a^2 + b^2 + c^2 (already given as 74\sqrt{74}).
  2. a2b2+b2c2+c2a2a^2b^2 + b^2c^2 + c^2a^2.

Step 2: Use the condition a+b+c=0a + b + c = 0

From a+b+c=0a + b + c = 0, the square of the sum is: (a+b+c)2=a2+b2+c2+2(ab+bc+ca)=0(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0 Simplify: a2+b2+c2=2(ab+bc+ca)a^2 + b^2 + c^2 = -2(ab + bc + ca) Substitute a2+b2+c2=74a^2 + b^2 + c^2 = \sqrt{74}: 74=2(ab+bc+ca)\sqrt{74} = -2(ab + bc + ca) Solve for ab+bc+caab + bc + ca: ab+bc+ca=742ab + bc + ca = -\frac{\sqrt{74}}{2}

Step 3: Calculate a2b2+b2c2+c2a2a^2b^2 + b^2c^2 + c^2a^2

We use the identity: a2b2+b2c2+c2a2=(ab+bc+ca)22abc(a+b+c)a^2b^2 + b^2c^2 + c^2a^2 = (ab + bc + ca)^2 - 2abc(a + b + c) Since a+b+c=0a + b + c = 0, the 2abc(a+b+c)2abc(a + b + c) term vanishes. Thus: a2b2+b2c2+c2a2=(ab+bc+ca)2a^2b^2 + b^2c^2 + c^2a^2 = (ab + bc + ca)^2 Substitute ab+bc+ca=742ab + bc + ca = -\frac{\sqrt{74}}{2}: (ab+bc+ca)2=(742)2=744=372(ab + bc + ca)^2 = \left(-\frac{\sqrt{74}}{2}\right)^2 = \frac{74}{4} = \frac{37}{2}

Step 4: Compute a4+b4+c4a^4 + b^4 + c^4

Using the formula from Step 1: a4+b4+c4=(a2+b2+c2)22(a2b2+b2c2+c2a2)a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2) Substitute a2+b2+c2=74a^2 + b^2 + c^2 = \sqrt{74} and a2b2+b2c2+c2a2=372a^2b^2 + b^2c^2 + c^2a^2 = \frac{37}{2}: a4+b4+c4=(74)22(372)a^4 + b^4 + c^4 = (\sqrt{74})^2 - 2\left(\frac{37}{2}\right) Simplify: a4+b4+c4=7437=37a^4 + b^4 + c^4 = 74 - 37 = 37

Final Answer:

a4+b4+c4=37a^4 + b^4 + c^4 = 37


Questions to Reflect On:

  1. Why does a+b+c=0a + b + c = 0 simplify the problem significantly?
  2. How is the symmetric sum a2b2+b2c2+c2a2a^2b^2 + b^2c^2 + c^2a^2 derived using basic algebraic identities?
  3. What role does the condition a2+b2+c2=74a^2 + b^2 + c^2 = \sqrt{74} play in this calculation?
  4. Can this method be generalized for other values of a+b+ca + b + c and a2+b2+c2a^2 + b^2 + c^2?
  5. What would the result be if a+b+c0a + b + c \neq 0?

Tip:

Always start by simplifying given conditions, as they often reduce the complexity of the problem through relationships among the variables.

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Math Problem Analysis

Mathematical Concepts

Algebra
Symmetric Functions
Polynomials

Formulas

a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2)
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
a^2b^2 + b^2c^2 + c^2a^2 = (ab + bc + ca)^2 - 2abc(a + b + c)

Theorems

Expansion of polynomial identities
Symmetric sum properties

Suitable Grade Level

Grades 10-12