Question

Find the value of

I)a^2 + b^2 when a + b = 7 and ab = 12

ii) x^2 + 1/x^2 and x^4 + 1/x^4 if x + 1/x = 4

Answer 1

$\huge{\texttt{\fcolorbox{Red}{aqua}{Hey Mate!!!}}}$

If a + b = 7
ab = 12

I) a^2+b^2
$(a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab \\ a {}^{2} + b {}^{2} = (a + b) {}^{2} - 2ab \\ a {}^{2} + b {}^{2} = 7 {}^{2} - 2 \times 12 \\ a {}^{2} + b {}^{2} = 49 - 24 \\ a {}^{2} + b {}^{2} = 25$
ii) x + 1/x = 4
$(x + \frac{1}{x} ) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } + 2 \times x \times \frac{1}{x} \\ x {}^{2} + \frac{1}{x {}^{2} } = (x + \frac{1}{x} ) {}^{2} - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 4 {}^{2} - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14$
Now. x^4 +1/x^4
$(x {}^{2} + \frac{1}{x} ) {}^{2} = x {}^{4} + \frac{1}{x {}^{4} } + 2 \times x {}^{2} \times \frac{1}{x {}^{2} } \\ x {}^{4} + \frac{1}{x {}^{4} } = (x {}^{2} + \frac{1}{x {}^{2} } ) {}^{2} - 2 \\ x {}^{4} + \frac{1}{x {}^{4} } = (14) {}^{2} - 2 \\ x {}^{4} + \frac{1}{x {}^{4} } = 196 - 2 \\ x {}^{4} + \frac{1}{x {}^{4} } = 194$
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Answer 2

Hey !!

(1) a + b = 7 and Ab = 12.

To find :- a² + b²

=> a² + b² = ( a + b )² - 2ab

=> ( 7 )² - 2 × 12

=> 49 - 24

=> 25

(2) X + 1/X = 4

Squaring both sides , we get

( X + 1/X )² = 4²

X² + 1/X² + 2 × X × 1/X = 16

X² + 1/X² + 2 = 16

X² + 1/X² = 14.

Squaring both sides , we get

( X² + 1/X²)² = 14²

X⁴ + 1/X⁴ + 2 × X² × 1/X² = 196

X⁴ + 1/X⁴ + 2 = 196

X⁴ + 1/X⁴ = 194.