If -x2-7- - 9 then find the values of x-

|x^2 7|≤ 9 9≤ x^2 7≤ 9 x^2 7≥ 9 and x^2 7≤ 9 ——————— x^2 7≥ 9 x^2≥ 2 #160; #160; True for all #34;x #34;——————— ...

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Modulus of a Complex Number

If |x2-7| ≤...

Question

If $| x^{2} - 7 | \leq 9$ then find the values of $x$ .

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Solution

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x + \sqrt{x^{2} - 9} = 4

x - \sqrt{x^{2} - 9}

x = \frac{5}{(2!) 3} + \frac{5 \cdot 7}{(3!) 3^{2}} + \frac{5 \cdot 7 \cdot 9}{(4!) 3^{3}} + . . . . .

x^{2} + 4 x

x - y = 7

x y = 9

(x^{2} + y^{2})

x - \frac{1}{x} = 7

x^{2} + \frac{1}{x^{2}} .

x^{2} - \frac{1}{9} = 0

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