Question

If \[\alpha \], \[\beta \] are the real and distinct roots of \[{x^2} + px + q = 0\] and \[{\alpha ^4}\], \[{\beta ^4}\] are the roots of \[{x^2} - rx + s = 0\], then the equation \[{x^2} - 4qx + 2{q^2} - r = 0\] has always
A) Two real roots
B) Two negative roots
C) Two positive roots
D) One positive and one negative root

Answer 1

Hint: Here, we will first use that in in an equation \[a{x^2} + bx + c\], the sum of roots of the equation is equal \[ - a\] and product is equal to \[b\] in both the given equations to find our own conditions. Then we will use the property \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] to find the required value.

Complete step by step solution: We are given that the real and distinct roots equation \[{x^2} + px + q = 0\] are \[\alpha \], \[\beta \] and \[{\alpha ^4}\], \[{\beta ^4}\] are the roots of \[{x^2} - rx + s = 0\].

Since we know that in an equation \[a{x^2} + bx + c\], the sum of roots of the equation is equal \[ - a\] and product is equal to \[b\].

Using this in the equation \[{x^2} + px + q = 0\], we get
\[ \Rightarrow \alpha + \beta = - p\]
 \[ \Rightarrow \alpha \beta = q\]

Now, considering the equation \[{x^2} - rx + s = 0\], we get
\[ \Rightarrow {\alpha ^4} + {\beta ^4} = r{\text{ ......eq.(1)}}\]
\[ \Rightarrow {\alpha ^4}{\beta ^4} = s{\text{ ......eq.(2)}}\]

Adding and subtracting the left side of the equation (1) by \[2{\alpha ^2}{\beta ^2}\], we get
\[ \Rightarrow {\alpha ^4} + {\beta ^4} + 2{\alpha ^2}{\beta ^2} - 2{\alpha ^2}{\beta ^2} = r\]

Using the property, \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] in the above equation, we get
\[ \Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2\alpha \beta = r\]

Adding and subtracting the left side of the above equation by \[2\alpha \beta \] in the bracket, we get
\[ \Rightarrow {\left( {{\alpha ^2} + {\beta ^2} + 2\alpha \beta - 2\alpha \beta } \right)^2} - 2{\alpha ^2}{\beta ^2} = r\]

Using the property, \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] in the above equation, we get
\[ \Rightarrow {\left( {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right)^2} - 2\alpha \beta = r\]

Substituting the value of \[\alpha + \beta \] and \[\alpha \beta \] in the above equation, we get
\[
   \Rightarrow {\left( {{{\left( { - p} \right)}^2} - 2q} \right)^2} - 2{q^2} = r \\
   \Rightarrow {\left( {{p^2} - 2q} \right)^2} - 2{q^2} = r \\
   \Rightarrow {p^4} + 4{q^2} - 4{p^2}q - 2{q^2} = r \\
   \Rightarrow {p^4} + 2{q^2} - 4{p^2}q = r \\
 \]

Rearranging the above equation, we get
\[
   \Rightarrow 2{q^2} - r = 4{p^2}q - {p^4} \\
   \Rightarrow 2{q^2} - r = {p^2}\left( {4q - {p^2}} \right) \\
 \]

Since we know that \[{p^2} > 0\] and \[4q - {p^2} < 0\], so the above equation will become
\[ \Rightarrow 2{q^2} - r < 0\]

Since a product of the roots can only be negative if one root is negative and one root is positive.

Using the above equation, we get that \[{x^2} - 4qx + 2{q^2} - r = 0\] has one positive and one negative root,

Hence, option D is correct.

Note: A quadratic is a type of problem that deals with a variable multiplied by itself and an operation known as squares. One should know that in an equation \[a{x^2} + bx + c\], the sum of roots of the equation is equal \[ - a\] and product is equal to \[b\]. We need to know the product of a negative number and a positive number is negative.