Question

Correct rationalization of 1/(7+5√2) is
a. 5 -7√2
b. 5√2-7
c. 7+5√2
d. None of these

Answer 1

Answer:

b) 5√2- 7 is the correct answer

Answer 2

$\huge\color{aqua}\ \boxed{\colorbox{black}{Answer : ♞ }}$

$\frac{ 1 }{ (7+5 \sqrt{ 2) } }$

$\color{red}SOLUTION:$

$Rationalize \: the \: denominator \: of \\ \frac{1}{7+5\sqrt{2}} \: by \: multiplying \: \\ numerator \: and \: denominator \\ by 7-5\sqrt{2}.$

$= > \frac{7-5\sqrt{2}}{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}$

$Consider \: \left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right). \\ Multiplication \: can \: be \: \\ transformed \: into \: difference \: \\ of squares \: using \: the \: rule: \\ \color{blue}\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.$

$= > \frac{7-5\sqrt{2}}{7^{2}-\left(5\sqrt{2}\right)^{2}}$

$Calculate \: 7 \: to \: the \: power \: of \\ 2 \: and \: get \: 49.$

$= > \frac{7-5\sqrt{2}}{49-\left(5\sqrt{2}\right)^{2}}$

$Expand \left(5\sqrt{2}\right)^{2}.$

$= > \frac{7-5\sqrt{2}}{49-5^{2}\left(\sqrt{2}\right)^{2}}$

$Calculate \: 5 \: to \: the \: power \: of \\ 2 \: and \: get \: 25.$

$= > \frac{7-5\sqrt{2}}{49-25\left(\sqrt{2}\right)^{2}}$

$The \: square \: of \: \sqrt{2} \: is \: 2.$

$= > \frac{7-5\sqrt{2}}{49-25\times 2}$

$Multiply \: 25 \: and \: 2 \: to \: get \: 50.$

$= > \frac{7-5\sqrt{2}}{49-50}$

$Subtract \: 50 \: from \: 49 \: to \: get -1.$

$= > \frac{7-5\sqrt{2}}{-1}$

$Anything \: divided \: by \: -1 \: gives \: \\ its \: opposite. \: To \: find \: the \\ opposite \: of \: 7-5\sqrt{2}, \: find \: the \\ opposite \: of \: each \: term.$

$= > -7-\left(-5\sqrt{2}\right)$

$The \: opposite \: of \: -5\sqrt{2} \: \: is \: \: 5\sqrt{2}.$

$= > \: -7+5\sqrt{2}$

$\color{blue}5\sqrt{2}-7\approx 0.071067812$