Question

please solve the right method
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Answer 1

Question 1:

Given :- x + y = 4 and xy = 2

Find the value of x² + y²

Solution 1:

We will use the identity (a+b)² = a² + b² +2ab

Squaring (x + y =4)

=> (x + y)² = (4)²

=> x² + y² +2xy = 16

=> x² + y² +2(2) = 16 [value of xy is given]

=> x² + y² + 4 = 16

=> x² + y² = 16-4

=> x² + y² = 12

Question 2:

If ${x}^{2} + \frac{1}{ {x}^{2} } = 18$,

find $x + \frac{1}{x} = {?}$

Answer 2:

Again we will use the same identity (above mentioned)

$\left(x + \frac{1}{x} \right) ^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \cancel{ x }\times \frac{1}{ \cancel{x}}$

We will put the value of x² + 1/x²

$= > \left(x + \frac{1}{x} \right)^{2} = 18 + 2$

$= > \left (x + \frac{1}{x} \right) ^{2} = 20$

$= > x + \frac{1}{x} = \sqrt{20}$

$= > x + \frac{1}{x} = 2 \sqrt{5}$

Question 3:

If $x + \frac{1}{x} = 12$,

find$x - \frac{1}{x} = {?}$

Answer 3:

Squaring x + 1/x = 12

$= > \left(x + \frac{1}{x} \right) {}^{2} = (12) {}^{2}$

$= > {x}^{2} + \frac{1}{ {x}^{2} } + 2( \cancel{x})( \frac{1}{ \cancel{x}} ) = 144$

$= > {x}^{2} + \frac{1}{ {x}^{2} } = 144 - 2$

${x}^{2} + \frac{1}{ {x}^{2} } = 142$

Now we will square $\left(x - \frac{1}{x} \right) {}^{2}$

$= > \left(x - \frac{1}{x} \right) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2$

x will cancel out

Substituting the value of ${x}^{2} + \frac{1}{ {x}^{2} }$

$= > \left(x - \frac{1}{x} \right) {}^{2} = 142 - 2$

$= > \left(x - \frac{1}{x} \right) {}^{2} = 140$

$= > x - \frac{1}{x} = \sqrt{140}$

$= > x - \frac{1}{x} = 2 \sqrt{35}$