# If $\alpha ,\beta$ are the roots of the equation ${x^2} - 4\sqrt 2 kx + 2{e^{4\ln k}} - 1 = 0$ for some k and ${\alpha ^2} + {\beta ^2} = 66$, then ${\alpha ^3} + {\beta ^3}$ is equal toA. $- 32\sqrt 2$B. $280\sqrt 2$C. $- 280\sqrt 2$D. $248\sqrt 2$

Verified
149.1k+ views
Hint: Comparing the given quadratic equation with the general quadratic equation, we find the values of a, b and c and substitute them in the formula for finding roots of quadratic equation. Assuming the roots as $\alpha ,\beta$ and then substituting them in the equation ${\alpha ^2} + {\beta ^2} = 66$ will give us the value of k, which when substituted back in equation of roots will give us the value of roots.
* General form of quadratic equation is $a{x^2} + bx + c = 0$ and the roots are given by the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

We have the quadratic equation ${x^2} - 4\sqrt 2 kx + 2{e^{4\ln k}} - 1 = 0$.
On comparing with general quadratic equation $a{x^2} + bx + c = 0$ we get,
$a = 1 \\ b = - 4\sqrt 2 k \\ c = 2{e^{4\ln k}} - 1 \\$
We can solve the value of c further by using the property of log : $m(\ln n) = \ln ({n^m})$
Here $m = 4,n = k$
$\Rightarrow 4\ln k = \ln {k^4}$
So, the value of c becomes
$c = 2{e^{\ln {k^4}}} - 1$
Since, log and exponent cancel each other, we get
$c = 2({k^4}) - 1$
Now substituting the values of a, b and c in formula to find the roots i.e. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow \dfrac{{ - ( - 4\sqrt 2 k) \pm \sqrt {{{( - 4\sqrt 2 k)}^2} - 4 \times 1 \times (2{k^4} - 1)} }}{{2 \times 1}} \\ \Rightarrow \dfrac{{4\sqrt 2 k \pm \sqrt {32{k^2} - 8{k^4} + 4} }}{2} \\$
Take common the value 4 inside the square root.
$\Rightarrow \dfrac{{4\sqrt 2 k \pm \sqrt {4(8{k^2} - 2{k^4} + 1)} }}{2}$
Bring outside the value of square root of 4 i.e. 2
$\Rightarrow \dfrac{{4\sqrt 2 k \pm 2\sqrt {8{k^2} - 2{k^4} + 1} }}{2}$
Cancel 2 from both numerator and denominator.
$\Rightarrow 2\sqrt 2 k \pm \sqrt {8{k^2} - 2{k^4} + 1}$
S, our two roots are
$\alpha = 2\sqrt 2 k + \sqrt {8{k^2} - 2{k^4} + 1} \\ \beta = 2\sqrt 2 k - \sqrt {8{k^2} - 2{k^4} + 1} \\$
Now we substitute the values of $\alpha ,\beta$ in equation ${\alpha ^2} + {\beta ^2} = 66$.
$\Rightarrow {\left( {2\sqrt 2 k + \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} + {\left( {2\sqrt 2 k - \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = 66$
Using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$ we solve LHS of equation.
The first term is solved as
$\Rightarrow {\left( {2\sqrt 2 k + \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = {(2\sqrt 2 k)^2} + {(\sqrt {8{k^2} - 2{k^4} + 1} )^2} + 2(2\sqrt 2 k)(\sqrt {8{k^2} - 2{k^4} + 1} ) \\ \Rightarrow {\left( {2\sqrt 2 k + \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = 8{k^2} + 8{k^2} - 2{k^4} + 1 + 4\sqrt 2 k\sqrt {8{k^2} - 2{k^4} + 1} \\ \Rightarrow {\left( {2\sqrt 2 k + \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = 16{k^2} - 2{k^4} + 1 + 4\sqrt 2 k\sqrt {8{k^2} - 2{k^4} + 1} \\$
And second term is solved as
$\Rightarrow {\left( {2\sqrt 2 k - \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = {(2\sqrt 2 k)^2} + {(\sqrt {8{k^2} - 2{k^4} + 1} )^2} - 2(2\sqrt 2 k)(\sqrt {8{k^2} - 2{k^4} + 1} ) \\ \Rightarrow {\left( {2\sqrt 2 k - \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = 8{k^2} + 8{k^2} - 2{k^4} + 1 - 4\sqrt 2 k\sqrt {8{k^2} - 2{k^4} + 1} \\ \Rightarrow {\left( {2\sqrt 2 k - \sqrt {8{k^2} - 2{k^4} + 1} } \right)^2} = 16{k^2} - 2{k^4} + 1 - 4\sqrt 2 k\sqrt {8{k^2} - 2{k^4} + 1} \\$
Adding both terms we get LHS of the equation
$\Rightarrow 16{k^2} - 2{k^4} + 1 + 4\sqrt 2 k\sqrt {8{k^2} - 2{k^4} + 1} + 16{k^2} - 2{k^4} + 1 - 4\sqrt 2 k\sqrt {8{k^2} - 2{k^4} + 1} \\ \Rightarrow 16{k^2} - 2{k^4} + 1 + 16{k^2} - 2{k^4} + 1 \\ \Rightarrow 32{k^2} - 4{k^4} + 2 \\$
Now equating LHS to RHS we get
$\Rightarrow 32{k^2} - 4{k^4} + 2 = 66$
Shifting all constants to one side of the equation
$\Rightarrow 32{k^2} - 4{k^4} = 66 - 2 \\ \Rightarrow 32{k^2} - 4{k^4} = 64 \\$
Take 4 common from terms on LHS of the equation
$\Rightarrow 4(8{k^2} - {k^4}) = 16 \times 4$
Cancel out same terms from both sides of the equation
$\Rightarrow 8{k^2} - {k^4} = 16 \\ \Rightarrow 8{k^2} - {k^4} - 16 = 0 \\$
Shift all the terms to one side of the equation.
$\Rightarrow {k^4} + 16 - 8{k^2} = 0$
This can be written as
$\Rightarrow {({k^2})^2} + {(4)^2} - 2({k^2})(4) = 0$
Comparing to the formula ${(a - b)^2} = {a^2} + {b^2} - 2ab$ we get ${({k^2})^2} + {(4)^2} - 2({k^2})(4) = ({k^2} - 4)$.
$\Rightarrow ({k^2} - 4) = 0 \\ \Rightarrow {k^2} = 4 \\$
Taking square root on both sides
$\Rightarrow \sqrt {{k^2}} = \sqrt 4$
Cancel out square root by square power
$\Rightarrow k = \pm 2$
Now substitute the value of k in the equation of roots.
Case 1: When $k = 2$
$\Rightarrow \alpha = 2\sqrt 2 \times 2 + \sqrt {8{{(2)}^2} - 2{{(2)}^4} + 1} \\ \Rightarrow \alpha = 4\sqrt 2 + \sqrt {32 - 32 + 1} \\ \Rightarrow \alpha = 4\sqrt 2 + \sqrt 1 \\ \Rightarrow \alpha = 4\sqrt 2 + 1 \\$
And
$\Rightarrow \beta = 2\sqrt 2 \times 2 - \sqrt {8{{(2)}^2} - 2{{(2)}^4} + 1} \\ \Rightarrow \beta = 4\sqrt 2 - \sqrt {32 - 32 + 1} \\ \Rightarrow \beta = 4\sqrt 2 - \sqrt 1 \\ \Rightarrow \beta = 4\sqrt 2 - 1 \\$
So, $\alpha = 4\sqrt 2 + 1,\beta = 4\sqrt 2 - 1$
Case 2: When $k = - 2$
$\Rightarrow \alpha = 2\sqrt 2 \times ( - 2) + \sqrt {8{{( - 2)}^2} - 2{{( - 2)}^4} + 1} \\ \Rightarrow \alpha = - 4\sqrt 2 + \sqrt {32 - 32 + 1} \\ \Rightarrow \alpha = - 4\sqrt 2 + \sqrt 1 \\ \Rightarrow \alpha = - 4\sqrt 2 + 1 \\$
And
$\Rightarrow \beta = 2\sqrt 2 \times ( - 2) - \sqrt {8{{(2)}^2} - 2{{(2)}^4} + 1} \\ \Rightarrow \beta = - 4\sqrt 2 - \sqrt {32 - 32 + 1} \\ \Rightarrow \beta = - 4\sqrt 2 - \sqrt 1 \\ \Rightarrow \beta = - 4\sqrt 2 - 1 \\$
So, $\alpha = - 4\sqrt 2 + 1,\beta = - 4\sqrt 2 - 1$
We see that if we take negative common from the values of case 2 then we get roots of case 1 with negative signs along them. So we can say that
$\alpha = - 4\sqrt 2 + 1 = - (4\sqrt 2 - 1) = - \beta \\ \beta = - 4\sqrt 2 - 1 = - (4\sqrt 2 + 1) = - \alpha \\$
Roots are the same in both cases.
Now we find the value of ${\alpha ^3} + {\beta ^3}$
We know the formula ${\alpha ^3} + {\beta ^3} = {(\alpha + \beta )^3} - 3\alpha \beta (\alpha + \beta )$
Substituting the value of $\alpha = 4\sqrt 2 + 1,\beta = 4\sqrt 2 - 1$we get
$\Rightarrow {\alpha ^3} + {\beta ^3} = {(4\sqrt 2 + 1 + 4\sqrt 2 - 1)^3} - 3(4\sqrt 2 + 1)(4\sqrt 2 - 1)(4\sqrt 2 + 1 + 4\sqrt 2 - 1)$
Since we know $(a + b)(a - b) = {a^2} - {b^2}$
$\Rightarrow {\alpha ^3} + {\beta ^3} = {(8\sqrt 2 )^3} - 3\{ {(4\sqrt 2 )^2} - {1^2})\} 8\sqrt 2 \\ \Rightarrow {\alpha ^3} + {\beta ^3} = 1024\sqrt 2 - 3(32 - 1)8\sqrt 2 \\$
Multiplying the terms in the bracket
$\Rightarrow {\alpha ^3} + {\beta ^3} = 1024\sqrt 2 - 3(31)8\sqrt 2 \\ \Rightarrow {\alpha ^3} + {\beta ^3} = 1024\sqrt 2 - 744\sqrt 2 \\ \Rightarrow {\alpha ^3} + {\beta ^3} = 280\sqrt 2 \\$

So, the correct answer is “Option B”.

Note: Students can many times make the mistake of not solving the log part in the beginning which will make all our calculations way much complex. Also, many students make the mistake of not changing the sign of the value when taking it to the opposite side of the equation.