Question

Solve the equation ${{\text{x}}^6}{\text{ - 18}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8 = 0}}$ , whose one of roots is $\sqrt 6 {\text{ - 2}}$.

Answer 1

- Hint: As the given root is irrational, it occurs in pairs. Therefore, another root is $- \sqrt 6 {\text{ - 2}}$. We will make a function with these two roots and solve the given problem.

Complete step-by-step solution -

Now, given roots are $\sqrt 6 {\text{ - 2}}$ and $- \sqrt 6 {\text{ - 2}}$. So, we can make a function.
Therefore, function is h (x) = ${\text{(x + 2 + }}\sqrt 6 ){\text{(x + 2 - }}\sqrt 6 )$
h (x) = ${{\text{x}}^2}{\text{ + 4x - 2}}$
Now, to find other roots, we have to divide f (x) = ${{\text{x}}^6}{\text{ - 18}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8}}$ by function h (x).
Now, by division theorem,
Dividend = divisor x quotient + remainder
As, the function h (x) is formed by the roots of f (x), so remainder = 0
f (x) = h(x). quotient
quotient = $\dfrac{{{{\text{x}}^6}{\text{ - 18}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8}}}}{{{{\text{x}}^2}{\text{ + 4x - 2}}}}$.

${{\text{x}}^2}{\text{ + 4x - 2}}\mathop{\left){\vphantom{1\begin{gathered} {{\text{x}}^6}{\text{ - 18}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8}} \\\ \underline {{{\text{x}}^6}{\text{ + 4}}{{\text{x}}^5}{\text{ - 2}}{{\text{x}}^4}} \\\ \- 4{{\text{x}}^5}{\text{ - 16}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + }}{\text{8}} \\\ \underline { - 4{{\text{x}}^5}{\text{ - 16}}{{\text{x}}^4}{\text{ + 8}}{{\text{x}}^3}} \\\ {\text{8}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8}} \\\ \underline {8{{\text{x}}^3}{\text{ + 32}}{{\text{x}}^2}{\text{ - 16x}}} \\\ {\text{4}}{{\text{x}}^2}{\text{ - 16x + 8}} \\\ \underline {{\text{4}}{{\text{x}}^2}{\text{ - 16x + 8}}} \\\ \underline 0 {\text{ }} \\\ \end{gathered} }}\right. \\!\\!\\!\\!\overline{\,\,\,\vphantom 1{\begin{gathered} {{\text{x}}^6}{\text{ - 18}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8}} \\\ \underline {{{\text{x}}^6}{\text{ + 4}}{{\text{x}}^5}{\text{ - 2}}{{\text{x}}^4}} \\\ \- 4{{\text{x}}^5}{\text{ - 16}}{{\text{x}}^4}{\text{ + 16}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + }}{\text{8}} \\\ \underline { - 4{{\text{x}}^5}{\text{ - 16}}{{\text{x}}^4}{\text{ + 8}}{{\text{x}}^3}} \\\ {\text{8}}{{\text{x}}^3}{\text{ + 28}}{{\text{x}}^2}{\text{ - 32x + 8}} \\\ \underline {8{{\text{x}}^3}{\text{ + 32}}{{\text{x}}^2}{\text{ - 16x}}} \\\ {\text{-4}}{{\text{x}}^2}{\text{ - 16x + 8}} \\\ \underline {{\text{-4}}{{\text{x}}^2}{\text{ - 16x + 8}}} \\\ \underline 0 {\text{ }} \\\ \end{gathered} }}} \limits^{\displaystyle \,\,\, {{{\text{x}}^4}{\text{ - 4}}{{\text{x}}^3}{\text{ + 8x - 4}}}}$
quotient = ${{\text{x}}^4}{\text{ - 4}}{{\text{x}}^3}{\text{ + 8x - 4}}$
Now, to find the roots of quotient, we will put quotient = 0
Therefore, we get
${{\text{x}}^4}{\text{ - 4}}{{\text{x}}^3}{\text{ + 8x - 4}}$ = 0
${\text{(}}{{\text{x}}^2}{\text{ }} - {\text{ 2)(}}{{\text{x}}^2}{\text{ - 4x + 2) = 0}}$
Solving $({{\text{x}}^2}{\text{ - 2) = 0}}$. We will use the property ${\text{(}}{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}{\text{) = (x + y)(x - y)}}$, we get
${\text{x = }} \pm \sqrt 2$
Solving ${{\text{x}}^2}{\text{ - 4x + 2 = 0}}$ with the help of Sridharacharya rule ${\text{x = }}\dfrac{{{\text{ - b }} \pm \sqrt {{{\text{b}}^2}{\text{ - 4ac}}} }}{{2{\text{a}}}}$, we get
${\text{x = 2}} \pm \sqrt 2$
So, the roots are $\pm \sqrt 2$, $2 \pm \sqrt 2$ and $- 2 \pm \sqrt 6$.

Note: When we come up with such types of questions, we will first make an equation with the roots given in the question. Now, to find the other roots, we have to divide the given equation by the equation formed. After it, we will apply the division theorem to find the quotient and the roots will be found by putting the equation of quotient equals to zero.