Easy identity

Algebra Level 2

$\large \dfrac{\sqrt{50}+7}{\sqrt{50}-7}+\dfrac{\sqrt{50}-7}{\sqrt{50}+7} = \ ?$

201 197 200 198 199

5 solutions

Ian Carlo
Aug 4, 2015

What I did was set $x=\sqrt{50}$ and $y=7.$ Then:

$\begin{aligned} & \frac{x+y}{x-y} + \frac{x-y}{x+y} \\ &= \frac{2x^2+2y^2}{x^2-y^2} \\ &= \frac{2(\sqrt{50})^2 + 2(7)^2}{(\sqrt{50})^2-(7)^2} \\ &= \frac{100+98}{1} \\ &=\boxed{198} \end{aligned}$

Moderator note:

Yes. Applying a simple substitution makes it simpler. Here's an upvote!

I love your approach!

Rico Lee - 4 years, 11 months ago

Very simplistic. Possibly the best solution among all of these!

Raakin Kabir - 4 years, 9 months ago

I love this technique!

Maygrens Macatangay - 4 years, 6 months ago

I don't get the second step, I'm lost, please help :-(

Simon Cochrane - 4 years, 4 months ago

Its similar to how you make the denominators the same, when adding fractions

e.g (1/(x +1)) + (4/(x+6)) = 1(x + 6) + 4(x + 1) (x + 1)(x + 6)

= x + 6 + 4x + 4 (x + 1)(x + 6)

= 5x + 10
(x + 1)(x + 6)

By the way this was not my working t was by this site https://revisionmaths.com/gcse-maths-revision/algebra/algebraic-fractions that fully explained to me how to these kind of questions and apply them on other topic. Hope this helps :)

Matthew Tioxon - 4 years, 3 months ago

I understand the concept of substituting the question into an x and y formula, each with their own value, I think that it is by far the best method

Matthew Tioxon - 4 years, 3 months ago

Khang Nguyen Thanh
Aug 2, 2015

We have:

$\dfrac{\sqrt{50}+7}{\sqrt{50}-7}=\dfrac{(\sqrt{50}+7)^2}{(\sqrt{50}-7)(\sqrt{50}+7)}=99+14\sqrt{50}\\ \dfrac{\sqrt{50}-7}{\sqrt{50}+7}=\dfrac{(\sqrt{50}-7)^2}{(\sqrt{50}-7)(\sqrt{50}+7)}=99-14\sqrt{50}$

So, $\dfrac{\sqrt{50}+7}{\sqrt{50}-7}+\dfrac{\sqrt{50}-7}{\sqrt{50}+7}=\boxed{198}$

sqrt(50)-7 = x;

sqrt(50)+7 =y;

=> xy = 1; x+y = 2sqrt(50)

=> x/y + y/x = (x^2 + y^2)/xy

= [(x+y)^2-2xy]/xy

= 200-2=198

Khang Hoàng Hồng - 5 years, 10 months ago

What is myophia ?

Arjun Suryakant Shahi - 5 years, 10 months ago

Nearsightedness is called myophia

Deepak Shakya - 5 years, 10 months ago

Yes but the spelling is: myopia

Fabiø Marinacci - 5 years, 5 months ago

So you $x^{2}$ the numerator? Then use the identity $(a+b)^{2} = a^{2}+2ab+b^{2}$ and $a^{2}-b^{2}=(a-b)(a+b)$

Adam Phúc Nguyễn - 5 years, 10 months ago

Sit Yew
Jun 29, 2016

let v= $\sqrt{50}+7$ , u= $\sqrt{50}-7$ and uv=50-49=1 then,

$\frac{v}{u}$ + $\frac{u}{v}$

= $\frac{v^{2}+u^2}{uv}$

= $\frac{(v+u)^2-2uv}{uv}$

= $(2\sqrt{50})^{2}$ -2

=4x50-2

=198

Creative solution

suzan khach - 3 years, 9 months ago

Gia Hoàng Phạm
Sep 22, 2018

$\frac{\sqrt{50}+7}{\sqrt{50}-7}+\frac{\sqrt{50}-7}{\sqrt{50}+7}=\frac{(\sqrt{50}+7)^2+(\sqrt{50}-7)^2}{50-49}=2 \times 50+2 \times 49=2(50+49)=2 \times 99=\boxed{\large{198}}$

Daniel Sugihantoro
Jun 29, 2016

I do this, yeah more complicated but not use too much Logic

$\frac { \sqrt { 50 } +7 }{ \sqrt { 50 } -7 } +\frac { \sqrt { 50 } -7 }{ \sqrt { 50 } +7 } \\ =\frac { { (\sqrt { 50 } +7) }^{ 2 }\quad +\quad { (\sqrt { 50 } -7) }^{ 2 } }{ { \sqrt { 50 } }^{ 2 }-{ 7 }^{ 2 } } \\ =\frac { 50+49+14\sqrt { 50 } +50+49-14\sqrt { 50 } }{ 50-49 } \\ =\frac { 99\times 2 }{ 1 } =198$

pretty simple!

Matthew Tioxon - 4 years, 3 months ago

I too done the same

Chaitanya Chaitu Chaitu - 3 years ago

Easy identity

5 solutions

Moderator note:

v u \frac{v}{u} u v ​ + u v \frac{u}{v} v u ​

= v 2 + u 2 u v \frac{v^{2}+u^2}{uv} u v v 2 + u 2 ​

= ( v + u ) 2 − 2 u v u v \frac{(v+u)^2-2uv}{uv} u v ( v + u ) 2 − 2 u v ​

= ( 2 50 ) 2 (2\sqrt{50})^{2} ( 2 5 0 ​ ) 2 -2

=4x50-2

=198

0 pending reports

$\frac{v}{u}$ + $\frac{u}{v}$

= $\frac{v^{2}+u^2}{uv}$

= $\frac{(v+u)^2-2uv}{uv}$

= $(2\sqrt{50})^{2}$ -2