Question

∆Challenge To ALL"BRAINLY STARS"
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Answer 1

$\huge\purple{\underline{\text{1st question}}}$

From the polynomial identity:-

$\large\boxed{(a+b)^3=a^3+3a^2b+3ab^2+b^3}$

One can derive:-

$\rightarrow\left(x+\dfrac{1}{x}\right)^3=x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}$

Since we know that:-

$\rightarrow x^3+\dfrac{1}{x^3}=110$

Let us assume that:-

$\rightarrow t=x+\dfrac{1}{x}\ \cdots(1)$

Then

$\rightarrow t^3=x^3+3(x+\dfrac{1}{x})+\dfrac{1}{x^3}\ \therefore t^3=110+3t\ \cdots(2)$

$\iff\red{\underline{t^3-3t-110=0}}$

By substituting $t=5$ on the equation,

$\rightarrow 125-15-110=0$

Now we know that it has $t-5$ as a factor.

$\rightarrow\red{\underline{(t-5)(t^2+5t+22)=0}}\ \therefore t=5\text{ or }t^2+5t+22=0$

(I am considering imaginary solutions too.)

We know that:-

$\large\boxed{(a+b)^{2}=a^2+2ab+b^2}$

For $t=5$:-

$\rightarrow\left(x+\dfrac{1}{x}\right)^2=x^2+2+\dfrac{1}{x^2}$

$\rightarrow25=x^2+2+\dfrac{1}{x^2}\ \therefore x^2+\dfrac{1}{x^2}=23$

$\rightarrow\left(x^2+\dfrac{1}{x^2}\right)^2=x^4+2+\dfrac{1}{x^4}$

$\rightarrow 529=x^4+2+\dfrac{1}{x^4}\ \therefore \red{\underline{x^4+\dfrac{1}{x^4}=527}\tiny{\text{//}}}$

For $t^2+5t+22=0$:-

$\rightarrow t^2=-5t-22\ \cdots(3)$

$\rightarrow x^2+2+\dfrac{1}{x^2}=-5t-22$

$\rightarrow x^2+\dfrac{1}{x^2}=-5t-24$

$\rightarrow \left(x^2+\dfrac{1}{x^2}\right)^2=(-5t-24)^2$

$\rightarrow x^4+2+\dfrac{1}{x^4}=25t^2+240t+576$

By substituting $(3)$:-

$\rightarrow x^4+\dfrac{1}{x^4}=25(-5t-22)+240t+574$

$\rightarrow x^4+\dfrac{1}{x^4}=-125t-550+240t+574$

$\rightarrow x^4+\dfrac{1}{x^4}=115t+24$

Solutions of $t^2+5t+22=0$ are $t=\dfrac{-5\pm3\sqrt{7}i}{2}$.

$\therefore x^4+\dfrac{1}{x^4}=\red{\underline{\dfrac{-527\pm345\sqrt{7}i}{2}}\tiny{\text{//}}}$

$\huge\purple{\underline{\text{2nd Question}}}$

Given that:-

$\rightarrow\begin{cases} & x=(25+10\sqrt{6})^\frac{1}{2} \\ & y=(25-10\sqrt{6})^\frac{1}{2} \end{cases}$

Then

$\rightarrow\begin{cases} & x^2=25+10\sqrt{6} \\ & xy=5 \\ & y^2=25-10\sqrt{6} \end{cases}$

Thus

$\rightarrow x^2+2xy+y^2=60\ \therefore \red{\underline{x+y=2\sqrt{15}}\tiny{\text{//}}}\text{ as }x>0\text{ and }y>0$

$\huge\purple{\underline{\text{3rd question}}}$

Given:-

$\rightarrow\dfrac{\sqrt{1+x}}{1+\sqrt{1+x}}+\dfrac{\sqrt{1-x}}{1-\sqrt{1-x}}$

Let us suppose:-

$\rightarrow\begin{cases} & A=\sqrt{1+x} \\ & B=\sqrt{1-x} \end{cases}$

Then

$\rightarrow\begin{cases} & A^2=1+x \\ & AB=\sqrt{1-x^2} \\ & B^2=1-x \end{cases}$

Thus

$\rightarrow A^2+2AB+B^2=2+2\sqrt{1-x^{2}}$

$\rightarrow A^2-2AB+B^2=2-2\sqrt{1-x^{2}}$

Substituting the given value, $x=\dfrac{\sqrt{3}}{2}$:-

$\rightarrow (A+B)^2=2+1=3\ \therefore\underline{A+B=\sqrt{3}}$

$\rightarrow(A-B)^2=2-1=1\ \therefore\underline{A-B=1}$

$\rightarrow AB=\sqrt{1-\dfrac{3}{4}}=\dfrac{1}{2}$

Given:-

$\rightarrow\dfrac{A}{1+A}+\dfrac{B}{1-B}$

$=\dfrac{A-AB+B+AB}{1-B+A-AB}$

$=\dfrac{A+B}{1+A-B-AB}$

By substitution:-

$\rightarrow\dfrac{(A+B)}{1+(A-B)-AB}$

$=\dfrac{\sqrt{3}}{1+1-\dfrac{1}{2}}$

$=\dfrac{\sqrt{3}}{\dfrac{3}{2}}$

$=\dfrac{2}{\sqrt{3}}$

$=\red{\underline{\dfrac{2\sqrt{3}}{3}}\tiny\text{//}}$

$\huge\purple{\underline{\text{4th question}}}$

Given equation:-

$\rightarrow4^x-4^{x-1}=24$

As we know that the subtraction of exponents result in the division

$\rightarrow4^x-\dfrac{4^x}{4}=24$

$\iff4^{x}\times(1-\dfrac{1}{4})=24$

$\iff4^x=24\times\dfrac{4}{3}$

$\iff4^x=32$

Hence

$\rightarrow2^{2x}=2^5$

$\iff 2x=5\ \therefore x=\dfrac{5}{2}$

Hence

$\rightarrow(2x)^x=5^{\frac{5}{2}}\ \red{\underline{\therefore(2x)^x=25\sqrt{5}}\tiny{\text{//}}}$

$\huge\purple{\underline{\text{5th question}}}$

Given equation:-

$\rightarrow x=\dfrac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$

By componendo-dividendo

$\rightarrow\dfrac{x+1}{x-1}=\sqrt{\dfrac{p+2q}{p-2q}}$

Squaring both sides

$\rightarrow\left(\dfrac{x+1}{x-1}\right)^2=\dfrac{p+2q}{p-2q}$

$\iff\dfrac{x^2+2x+1}{x^2-2x+1}=\dfrac{p+2q}{p-2q}$

$\iff(x^2+2x+1)(p-2q)=(x^2-2x+1)(p+2q)$

$\iff px^2+2px+p-2qx^2-4qx-2q=px^2-2px+p+2qx^2-4qx+2q$

$\iff(p-2q)x^2-(p+2q)x^2+(2p-4q)x-(-2p-4q)x+(p-2q)-(p+2q)=0$

$\iff-4qx^2+4px-4q=0\ \therefore \red{\underline{qx^2-px+q=0}\tiny{\text{//}}}$