Question

1) If (n²+1)t=u + v and (n+1/n)t= u-v, prove that u/v = n+1/n-1 and find t in terms of u,v only

2) For what values of c is x=-3 a root of the equation, x²+(c-1)x-2=c²? What are then the other values of x?

Answer 1

Solution. 1) Prove that u/v = n+1/n-1.

According to the condition of the problem (n²+1)t=u + v and (n+1/n)t= u-v. Let's add u + v and u-v get

$$(n^2+1)t+(n+\frac{1}{n})t=u+v+u-v$$

$$(n^2+1)t+(\frac{n^2+1}{n})t=2u$$

$$u=\frac{(n^2+1)(n+1)t}{2n}$$

Subtract u-v from u + v get

$$u + v -(u -v)=(n^2+1)t-(n+\frac{1}{n})t$$

$$2v=(n^2+1)t-(\frac{n^2+1}{n})t$$

$$v=\frac{(n^2+1)(n-1)t}{2n}$$

Find the ratio u/v

$$\frac{u}{v}=\frac{\frac{(n^2+1)(n+1)t}{2n}}{\frac{(n^2+1)(n-1)t}{2n}}=\frac{n+1}{n-1}.$$

Find t in terms of u,v. 

$$(n+\frac{1}{n})t=u-v \to \frac{(n^2+1)t}{n}=u-v$$

On the other hand (n²+1)t=u + v. As result get

$$\frac {u+v}{n} = u - v \to n =\frac {u+v}{u-v}$$

Substituting n=(u+v)/(u-v) get

$$(\frac{(u+v)^2}{(u-v)^2}+1)t=u+v$$

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

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

2) If x=-3 a root of the equation, x²+(c-1)x-2=c² get

$$(-3)^2+(c-1)(-3)-2=c^2$$

$$c^2+3c-10=0$$

The roots of the quadratic equation

$$с_1=2$$

and

$$c_2=-5.$$

For c=2 get equation

$$x^2+(2-1)x-2=2^2 \to x^2+x-6=0.$$

The roots of the quadratic equation $x_1=-3$ and $x_2=2$ .

For c=-5 get equation

$$x^2+(-5-1)x-2=(-5)^2 \to x^2-6x-27=0.$$

The roots of the quadratic equation $x_3=-3$ and $x_4=9$.

The other values of x=2 and x =9

Answer. 1) $t=\frac{(u+v)(u-v)^2}{2(u^2+v^2)}$ ; 2) $с_1=2$ and $c_2=-5$; the other values of x=2 and x =9.