Question

Solve the following equations:
$10(x^4+1)-63x(x^2-1)+52x^2=0$.

Answer 1

$10(x^4+1)-63x(x^2-1)+52x^2=0$

$10x^4-63x^3+52x^2+63x+10=0$

Replace $x$ by $-\frac{1}{x}$

$\frac{10}{x^4}+\frac{63}{x^3}+\frac{52}{x^2}-\frac{63}{x}+10=0$

$\frac{10+63x+52x^2-63x^3+10x^4}{x^4}=0$

$10+63x+52x^2-63x^3+10x^4=0$

On replacing $x$ by $-\frac{1}{x}$ the equation become same , so the roots of the equation are of form

$\alpha ,-\frac{1}{\alpha },\beta ,-\frac{1}{\beta }$

$\alpha +(-\frac{1}{\alpha })+\beta +(-\frac{1}{\beta })=-\frac{b}{a}$

$\alpha -\frac{1}{\alpha }+\beta -\frac{1}{\beta }=-\frac{-63}{10}=\frac{63}{10}$ ........(i)

$\alpha .(-\frac{1}{\alpha })+\alpha \beta +\alpha (-\frac{1}{\beta })+\beta (-\frac{1}{\alpha })+(-\frac{1}{\alpha })(-\frac{1}{\beta })+\beta (-\frac{1}{\beta })=\frac{c}{a}$

$-1+\alpha \beta -\frac{\alpha }{\beta }-\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }-1=\frac{52}{10}$

$\alpha (\beta -\frac{1}{\beta })-\frac{1}{\alpha }(\beta -\frac{1}{\beta })=\frac{52}{10}+2$

$(\alpha -\frac{1}{\alpha })(\beta -\frac{1}{\beta })=\frac{72}{10}...........\left(ii\right)$

using $\left(i\right)$

$(\alpha -\frac{1}{\alpha })(\frac{63}{10}-(\alpha -\frac{1}{\alpha }))=\frac{72}{10}$

Put $(\alpha -\frac{1}{\alpha })=t$

$t(\frac{63}{10}-t)=\frac{72}{10}$

$63t-10t^2=72$

$10t^2-63t+72=0$

$t=\frac{63\pm \sqrt{-63\times -63-4\left(10\right)72}}{20}=\frac{63\pm 33}{20}$

$\Rightarrow t=\frac{24}{5},\frac{3}{2}$

(1) $t=\frac{3}{2}$

$\alpha -\frac{1}{\alpha }=\frac{3}{2}$

$2{\alpha }^2-2=3\alpha$

$2{\alpha }^2-3\alpha -2=02{\alpha }^2-4\alpha +\alpha -2=0$

$2\alpha (\alpha -2)+1(\alpha -2)=0$

$(2\alpha +1)(\alpha -2)=0$

$\alpha =2,-\frac{1}{2}$

(2) $t=\frac{24}{5}$

$\alpha -\frac{1}{\alpha }=\frac{24}{5}$

$5{\alpha }^2-5=24\alpha 5{\alpha }^2-24\alpha -5=05{\alpha }^2-25\alpha +\alpha -5=05\alpha (\alpha -5)+1(\alpha -5)=0(5\alpha +1)(\alpha -5)=0\alpha =5,-\frac{1}{5}$

So, the values of $x$ are $2,-\frac{1}{2},5,-\frac{1}{5}$.