Problem 3: Symmetrical Properties of Roots

Let one root be $a$ , and the other root be $2a$ .

By vieta's formulae we get $3a=2k+4 --(1)$ $2a^2=k^2+3k+2 --(2)$ hence $\displaystyle a=\frac{2(k+2)}{3}$ .

Subsituting in (2) we get $2\left(\frac{4(k+2)^2}{9}\right)=k^2+3k+2$ $\Rightarrow 8k^2+32k+32=9k^2+27k+18$ $\Rightarrow k^2-5k-14$ $\Rightarrow k=\frac{5\pm\sqrt{25+56}}{2}$ $\Rightarrow k= 7 or -2$ As $k=-2$ makes both roots equal to zero , we get our answer as $\boxed{7}$

Starting from equation's $(1)$ and $(2)$ from Shriram Lokhande's answer:

$3a=2k+4\\ { 3a=2(k+2)\longrightarrow (A) } \\$

$\\ 2{ a }^{ 2 }={ k }^{ 2 }+3k+2\\ { 2{ a }^{ 2 }=(k+2)(k+1)\longrightarrow (B) }$

squaring both sides of $(A)$ you get:

${ (3a) }^{ 2 }={ (2(k+2)) }^{ 2 }\\ 9{ a }^{ 2 }={ 4(k+2) }^{ 2 }\longrightarrow (C)$

divide $(C)$ by $(B)$ and solve:

$\quad \quad (C):\quad 9{ a }^{ 2 }={ 4(k+2) }^{ 2 }\\ \underline { \div \quad (B):\quad 2{ a }^{ 2 }=(k+2)(k+1) } \\ \quad \quad \quad \quad \quad \quad \frac { 9 }{ 2 } =4\frac { k+2 }{ k+1 } \\ \quad \quad \quad \quad \quad \quad \frac { 9 }{ 8 } =\frac { k+2 }{ k+1 }$

Now it's pretty obvious that $\boxed { k=7 }$

One root is R and the second root is 2R. Product of roots ((2R)(R)) is k^2+3k+2 and the sum (2R+R) is 2k+4. Equate both equations to R= and solve for k. There's 2 answer: 7 and -2, but only 7 works. PS: also please tell me how to write equations in Latex form. Thanks.