Question

Determine rational numbers p and q if 7+√5/7-√5 - 7-√5/7+√5=p-7√5 9.​

Answer 1

Correct question -

Determine rational numbers p and q if

$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7 \sqrt{5} q$

Solution -

$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7 \sqrt{5} q$

By rationalising the denominator we will get -

$\frac{(7 + \sqrt{5} )(7 + \sqrt{5}) }{(7 - \sqrt{5})(7 + \sqrt{5} ) } - \frac{(7 - \sqrt{5} )(7 - \sqrt{5} )}{(7 + \sqrt{5})(7 - \sqrt{5} )} = p - 7 \sqrt{5} q$

$\frac{(7 + \sqrt{5}) {}^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } - \frac{(7 - \sqrt{5}) {}^{2} }{ {(7)}^{2} - {( \sqrt{5}) }^{2} } = p - 7 \sqrt{5} q$

$\frac{ {(7)}^{2} + {( \sqrt{5} )}^{2} + 2 \times 7 \times \sqrt{5} }{49 - 5} - \frac{ {(7)}^{2} + {( \sqrt{5}) }^{2} - 2 \times 7 \times \sqrt{5} }{49 - 5} = p - 7 \sqrt{5} q$

$\frac{49 + 5 + 14 \sqrt{5} }{44} - \frac{49 + 5 - 14 \sqrt{5} }{44} = p - 7 \sqrt{5} q$

$\frac{54 + 14 \sqrt{5} }{44} - \frac{54 - 14 \sqrt{5} }{44} = p - 7 \sqrt{5} q$

$\frac{54 + 14 \sqrt{5} - (54 - 14 \sqrt{5} )}{44} = p - 7 \sqrt{5} q$

$\frac{54 + 14 \sqrt{5} - 54 + 14 \sqrt{5} }{44} = p - 7 \sqrt{5} q$

$\frac{28 \sqrt{5} }{44} = p - 7 \sqrt{5} q$

$\frac{7 \sqrt{5} }{11} = p - 7 \sqrt{5} q$

$7 \sqrt{5} \times \frac{1}{11} = p - 7 \sqrt{5} q$

We know that,

Rational number is always equal to other rational numbers and irrational numbers is always equal to other irrational numbers.

$p = \frac{1}{11}$

And,

$7 \sqrt{5} = - 7 \sqrt{5} q \\ q = \frac{7 \sqrt{5} }{ - 7 \sqrt{5} } \\ q = - 1$

Therefore the value of p is 1/11 and q is - 1

Answer 2

Step-by-step explanation:

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