Question

If a + b + c = 14, ab + bc + ac = 61,find a2 + b2 + c2

Answer 1

a + b + c = 14 ... (i)

ab + bc + ac = 61 ... (ii)

(a + b + c)² = (14)² ... (from i)

a² + b² + c² + 2(ab + bc + ac) = 196

a² + b² + c² + 2(61) = 196 ... (from ii)

a² + b² + c² + 122 = 196

a² + b² + c² = 196 - 122

a² + b² + c² = 74

Answer 2

$\bf \red{{a}^{2} + {b}^{2} + {c}^{2} = 74}$.

Given:

• $a + b + c = 14 \: ...eq1$
• $ab + bc + ac = 61...eq2$

To find:

• Find the value of ${a}^{2} + {b}^{2} + {c}^{2}$

Solution:

Identity to be used:

$\bf ( {x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz$

Step 1:

Squaring both sides of eq1.

$( {a + b + c)}^{2} = ( {14)}^{2}$

or

Expand identity in LHS

${a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac = 196$

or

${a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac) = 196$

Step 2:

Put the value from eq2.

${a}^{2} + {b}^{2} + {c}^{2} + 2(61) = 196$

or

${a}^{2} + {b}^{2} + {c}^{2} = 196 - 122$

or

${a}^{2} + {b}^{2} + {c}^{2} = 74$

Thus,

Value is $\bf \: {a}^{2} + {b}^{2} + {c}^{2} = 74$.

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