How to solve equations of this type

Question

My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:

Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?

The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.

Q2. If: $\frac{x}{x + y} = 5$ then what is $\frac{y}{x + y}$ equal to?

The answer for this is -5, I also don't know how.

Q3. If: $x^4 + y^4 = 6 x^2 y^2 \land x\neq y$ then what is $\frac{x^2 + y^2}{x^2 - y^2}$ equal to?

I guess the answer for this was $\sqrt{2}$ but I'm not sure.

Any body can explain how to solve these questions, and questions of the same pattern?

@LordSharktheUnknown this solves the second one, thanks. But what about the others. — Hatchi Roku, Dec 22 '18 at 15:56
The first one: $(x - \frac{1}{x})^2=x^2-2+\frac{1}{x^2} = 9$ — Cheong Sik Feng, Dec 22 '18 at 16:21

Rebellos · Accepted Answer · 2018-12-22T15:58:20.240

5

For the first one :

$$\left(x-\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \Leftrightarrow x^2 + \frac{1}{x^2} = \left(x-\frac{1}{x}\right)^2 + 2 \implies x^2+ \frac{1}{x^2} = 11$$

For the second one, observe that :

$$\frac{x}{x+y} + \frac{y}{x+y} = 1 \Rightarrow 5 + \frac{y}{x+y} = 1 \Leftrightarrow \frac{y}{x+y} = -4 $$

For the third one, a small hint :

$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$

Alternativelly, observe that it also is :

$$x^4 + y^4 = 6 x^2 y^2 \Leftrightarrow \frac{x^2}{y^2} + \frac{y^2}{x^2} = 6$$

edited Dec 22 '18 at 15:58

answered Dec 22 '18 at 15:56

Rebellos

21,324

@ÍgjøgnumMeg Thanks. You may as well edit to correct those. – Rebellos Dec 22 '18 at 15:59
Thanks very much, I guess I'll get A+ tomorrow. – Hatchi Roku Dec 22 '18 at 16:15
2

@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice. – Rebellos Dec 22 '18 at 18:25
thanks for the advice. – Hatchi Roku Dec 22 '18 at 20:10

score 1 · Answer 2 · answered Dec 22 '18 at 15:57

Hint:
1: $x - \dfrac{1}{x} = 3 \implies x^2 + \dfrac{1}{x^2} - 2 = 9$
Now use, $\big(x + \dfrac{1}{x}\big)^2 = x^2 + \dfrac{1}{x^2} + 2$

2: $\dfrac{x}{x+y} = \dfrac{1}{1 + \dfrac{y}{x}}$

3: $x^4 + y^4 = 6 * x^2 * y^2 \implies \dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression

score 1 · Answer 3 · answered Dec 22 '18 at 15:58

1

Notice that

$$x^2 + \frac{1}{x^2} = \left(x -\frac{1}{x}\right)^2 + 2 = 3^2 + 2 = 11$$

2

Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.

3

You can transform the equation and solve for $\frac{y}{x}$ for example, and find everything you need.

score 1 · Answer 4 · answered Dec 22 '18 at 16:08

1

Hints.

$1) x-\frac1x=3\implies x^2+\frac1{x^2}-2=9\\2)\frac y{x+y}+\frac x{x+y}=1\\3)x^4+y^4=6x^2y^2\\\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\\\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$

Divide the two and take the square root.

answered Dec 22 '18 at 16:08

Shubham Johri

17,659

Thanks for the detailed explanation of the last one. – Hatchi Roku Dec 22 '18 at 16:16

How to solve equations of this type

4 Answers4