Question

$if \: {a}^{2} + {b}^{2} = 13 \: and \: ab = 6...then \: find \: the \: value \: of \\ (i)a + b \\ (ii)a - b$
help me guys!! tomorrow's my exm and I'm not so well prepared....​

Answer 1

Answer:

a)a+b=5✔✔

b)a-b=1✔✔

Step-by-step explanation:

${a}^{2} + {b}^{2} = 13 \\ ({a + b})^{2} - 2ab = 13 \: \: \: \: \: ({a}^{2} + {b}^{2} = ({a + b})^{2} - 2ab) \\ ( {a + b})^{2} - 2 \times 6 = 13 \\ ( {a + b})^{2} = 13 + 12 \\ ( {a + b})^{2} = 25 \\ a + b = \sqrt{25 } \\ a + b = 5 \\ again \\ {a}^{2} + {b}^{2} = 13 \\ ( {a - b})^{2} + 2ab = 13 \: \: \ ({a}^{2} + {b}^{2} = ( {a - b})^{2} + 2ab) \\ ( {a - b})^{2} + 2 \times 6 = 13 \\ ( {a - b})^{2} + 12 = 13 \\ ( {a - b})^{2} = 13 - 12 \\ a - b = \sqrt{1} \\ a - b = 1$

I HOPE IT HELP YOU

Answer 2

Answer:

a)a+b=5✔✔

b)a-b=1✔✔

Step-by-step explanation:

\begin{lgathered}{a}^{2} + {b}^{2} = 13 \\ ({a + b})^{2} - 2ab = 13 \: \: \: \: \: ({a}^{2} + {b}^{2} = ({a + b})^{2} - 2ab) \\ ( {a + b})^{2} - 2 \times 6 = 13 \\ ( {a + b})^{2} = 13 + 12 \\ ( {a + b})^{2} = 25 \\ a + b = \sqrt{25 } \\ a + b = 5 \\ again \\ {a}^{2} + {b}^{2} = 13 \\ ( {a - b})^{2} + 2ab = 13 \: \: \ ({a}^{2} + {b}^{2} = ( {a - b})^{2} + 2ab) \\ ( {a - b})^{2} + 2 \times 6 = 13 \\ ( {a - b})^{2} + 12 = 13 \\ ( {a - b})^{2} = 13 - 12 \\ a - b = \sqrt{1} \\ a - b = 1\end{lgathered}

a

2

+b

2

=13

(a+b)

2

−2ab=13(a

2

+b

2

=(a+b)

2

−2ab)

(a+b)

2

−2×6=13

(a+b)

2

=13+12

(a+b)

2

=25

a+b=

25

a+b=5

again

a

2

+b

2

=13

(a−b)

2

+2ab=13 (a

2

+b

2

=(a−b)

2

+2ab)

(a−b)

2

+2×6=13

(a−b)

2

+12=13

(a−b)

2

=13−12

a−b=

1

a−b=1

I HOPE IT HELP YOU