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If a + b = 5 and a - b = 3, then the value of (a² + b²) is?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + b = 5 \\ \Leftrightarrow a^{2} + b^{2} + 2 a b = 25 . . . . . (i) \\ a - b = 3 \\ \Leftrightarrow a^{2} + b^{2} - 2 a b = 9 . . . . . (i i) \\ From equation (i) and (ii) \\ \Leftrightarrow 2 (a^{2} + b^{2}) = 34 \\ \Leftrightarrow a^{2} + b^{2} = 17 \end{aligned}

$\eqalign{ & a + b = 5{\text{ }} \cr & \Leftrightarrow {a^2} + {b^2} + 2ab = 25\,.....(i) \cr & a - b = 3 \cr & \Leftrightarrow {a^2} + {b^2} - 2ab = 9\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Leftrightarrow {\text{2}}\left( {{a^2} + {b^2}} \right) = 34 \cr & \Leftrightarrow {a^2} + {b^2} = 17 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a + b = 5 \\ \Leftrightarrow a^{2} + b^{2} + 2 a b = 25 . . . . . (i) \\ a - b = 3 \\ \Leftrightarrow a^{2} + b^{2} - 2 a b = 9 . . . . . (i i) \\ From equation (i) and (ii) \\ \Leftrightarrow 2 (a^{2} + b^{2}) = 34 \\ \Leftrightarrow a^{2} + b^{2} = 17 \end{aligned}

$\eqalign{ & a + b = 5{\text{ }} \cr & \Leftrightarrow {a^2} + {b^2} + 2ab = 25\,.....(i) \cr & a - b = 3 \cr & \Leftrightarrow {a^2} + {b^2} - 2ab = 9\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Leftrightarrow {\text{2}}\left( {{a^2} + {b^2}} \right) = 34 \cr & \Leftrightarrow {a^2} + {b^2} = 17 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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