Question

The quadratic equation with rational, coefficient the sum of the squares of whose root is 40 and the sum of the cubes of whose root is 208 is
A. \[{{x}^{2}}+4x+12=0\]
B. \[{{x}^{2}}-4x-12=0\]
C. \[{{x}^{2}}-4x+12=0\]
D. \[{{x}^{2}}+4x-12=0\]

Answer 1

Hint: In this question, we have given four equations. We have to find an equation which satisfies the conditions given in the equation. To find out that equation, we make the equations from the conditions given in the question and by solving those equations with the help of formulas, we get our desirable answer.

Complete step by step solution: 
Given are the four quadratic equations.
We have to find out the equation whose sum of the square of root is 40 and sum of the cubes of root is 208.
Let $\alpha ,\beta $be its roots.
That means ${{\alpha }^{2}}+{{\beta }^{2}}=40$ and ${{\alpha }^{3}}+{{\beta }^{3}}=208$ (given in the equation)
Now suppose $\alpha +\beta =m,\alpha \beta =n$
${{\alpha }^{2}}+{{\beta }^{2}}=40$
And we know formula of $({{x}^{3}}+{{y}^{3}})=(x+y)({{x}^{2}}-xy+{{y}^{2}})$
By putting x = $\alpha $and y = $\beta $ in the above equation, we get
$({{\alpha }^{3}}+{{\beta }^{3}})=(\alpha +\beta )({{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}})$ ${{x}^{2}}-4x-12=0$
$(\alpha +\beta )({{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}})=208$
Now put $(\alpha +\beta )=m$and $\alpha \beta =n$in the above equation, we get
$m({{\alpha }^{2}}+{{\beta }^{2}}-n)=208$
As given in the question ${{\alpha }^{2}}+{{\beta }^{2}}=40$, we put this value in the above equation, we get
$m(40-n)=208$ . . . . . . . .. . . . . . . . . . . . . . . . .(1)
We know ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
By putting $a=\alpha $ and $b=\beta $, we get
${{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $
By putting value of m and n in the above equation, we get
${{m}^{2}}=40+2n$ ( as ${{\alpha }^{2}}+{{\beta }^{2}}=40$)
From this, we get $n=\frac{{{m}^{2}}-40}{2}$
Put this value of n in equation (1) , we get
$m\left[ \left( 40 \right)-\left\{ \frac{{{m}^{2}}-40}{2} \right\} \right]=208$
By solving this equation, we get
${{m}^{3}}-120m+216=0$
By using long division to solve the above equation, we get
$(m-4)({{m}^{2}}+4m-104)=0$
By solving, we get (m-4) =0
$m=4$
Negative value is not possible.
So $m=4,n=-12$
From there, we get a quadratic equation ${{x}^{2}}-4x-12=0$
Option (B) is correct.

Note: To solve these types of questions, we must know from which method we are able to get the answer easily. Sometimes we get answers from the options easily. We have to take care while putting the formulas in the equation then we get our answer.