Tag Archives: Algebra
${x^{{x^2}}} = {x^{4x + 5}}$
The number of solution(s) the equation ${x^{{x^2}}} = {x^{4x + 5}}$ has:
(A) One solution only (B) Two solutions
(C) Three solutions (D) No solution
Solve for $x \in \mathbb{R}$ ${(6x + 5)^2} = \frac{{35}}{{3{x^2} + 5x + 2}}$
We have, $36{x^2} + 60x + 25 = \frac{{35}}{{3{x^2} + 5x + 2}}$
$ \Rightarrow 12(3{x^2} + 5x) + 25 = \frac{{35}}{{(3{x^2} + 5x) + 2}}$
Let, $3{x^2} + 5x = t$
So, $12t + 25 = \frac{{35}}{{t + 2}}$
$ \Rightarrow 12{t^2} + 49t + 15 = 0$ Continue reading Solve for $x \in \mathbb{R}$
${(6x + 5)^2} = \frac{{35}}{{3{x^2} + 5x + 2}}$
If $x+\frac {1}{x}=2,x\neq 0$$x^{99}+\frac {1}{x^{99}}=?$
We have, $x^2+1-2x=0$
$\Rightarrow (x-1)^2=0$
$\Rightarrow x=1$
$\therefore x^{99}+\frac {1}{x^{99}}=1+1=2$
$x^2-x-1=0$$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = ?$
$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = \frac{{({x^2} – 1)({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}}$
$\because {x^2} – x – 1 = 0,{x^2} – 1 = x$
$R(x) = \frac{{x({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}} = \frac{{({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^6}(x + 2)}}$
$\because {x^2} – x – 1 = 0,{x^2} = x + 1$ Continue reading $x^2-x-1=0$
$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = ?$
$(x+5)(x+6)(x+7)=504$
Which of the following is/are correct regarding solution to the equation $(x+5)(x+6)(x+7)=504$?
(A) One real solution
(B) Two complex solutions
(C) Three real solutions
(D) No real solution
Continue reading $(x+5)(x+6)(x+7)=504$