Tag Archives: Equation
Solve for x & y$\sqrt x + y = 7$$x + \sqrt y = 11$
${x^{{x^2}}} = {x^{4x + 5}}$
The number of solution(s) the equation ${x^{{x^2}}} = {x^{4x + 5}}$ has:
(A) One solution only (B) Two solutions
(C) Three solutions (D) No solution
Solve for $x \in \mathbb{R}$$7x^5 + x + 8 = 0$
-1 satisfies the given equation. Thus,
$7x^4 (x+1) – 7x^3 (x+1) + 7x^2 (x+1) -7x(x+1)+8(x+1)=0$
$\Rightarrow (x+1) (7x^4 – 7x^3 +7x^2 -7x +8) = 0$
Consider, $7x^4 – 7x^3 +7x^2 -7x +8 = 0$
Or, $7(x^4 – x^3 +x^2 -x) +8 = 0$ Continue reading Solve for $x \in \mathbb{R}$
$7x^5 + x + 8 = 0$
Solve for x, ${4^x} – {3^{x – \frac{1}{2}}} = {3^{x + \frac{1}{2}}} – {2^{2\left( {x – \frac{1}{2}} \right)}}$
We have, ${2^{2x}} + \frac{{{2^{2x}}}}{2} = {3^x}\sqrt 3 + \frac{{{3^x}}}{{\sqrt 3 }}$
$ \Rightarrow \frac{3}{2}{.2^{2x}} = \frac{4}{{\sqrt 3 }}{.3^x}$
$ \Rightarrow \frac{{{2^{2x}}}}{8} = \frac{{{3^x}}}{{3\sqrt 3 }}$
$ \Rightarrow {2^{2x – 3}} = {3^{x – \frac{3}{2}}}$
$ \Rightarrow {2^{2\left( {x – \frac{3}{2}} \right)}} = {3^{x – \frac{3}{2}}}$
$ \Rightarrow {4^{x – \frac{3}{2}}} = {3^{x – \frac{3}{2}}}$
The above is only possible if $x – \frac{3}{2} = 0$ or $x=\frac {3}{2}$.
Solve the equation, $x^3+3x+2i=0$Where, $i=\sqrt {-1}$
$x=-i$ satisfies the given equation,
$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$
$\Rightarrow (x+i)(x^2-xi+2)=0$
Now consider, $x^2-xi+2=0$
$x=-i$ again satisfies the above equation,
$\therefore x(x+i)-2i(x+i)=0$
$\Rightarrow (x+i)(x-2i)=0$
$x=2i$ is another solution,
So, $x=-i$, $x=-i$, $x=2i$ are the three solutions.
Number of solutions to the equation ${x^{{x^x}}} = x$
(A) 1 (B) 2 (C) 3 (D) None of the options given
We have, ${x^{\left( {{x^x}} \right)}} = x$
Taking log, ${x^x}\ln |x| = \ln |x|$
$ \Rightarrow \ln |x|\left( {{x^x} – 1} \right) = 0$
$\ln |x| = 0,x = \pm 1$ Or ${x^x} = 1$
Taking log, $x\ln |x| = 0$
x = 0 Or $\ln |x| = 0,x = \pm 1$
x = 0 is rejected as it leads to $0^0$ situation.
Hence, two solutions or option (B).