Question

If 3 and $1 + \sqrt 2 $ are two roots of a cubic equation with rational coefficients , then the equation is
A) ${x^3} - 5{x^2} + 9x - 9 = 0$
B) ${x^3} - 3{x^2} - 4x + 12 = 0$
C) ${x^3} - 5{x^2} + 7x - 3 = 0$
D) None of these

Answer 1

Hint:
we are given two roots and since $1 + \sqrt 2 $is irrational and we know that irrational roots occur in pairs we get the other root to be $1 - \sqrt 2 $and now taking these roots to be a , b and c respectively we get the required cubic equation by ${x^3} - \left( {a + b + c} \right){x^2} + \left( {ab + bc + ca} \right)x - abc = 0$

Complete step by step solution:
We are given that 3 and $1 + \sqrt 2 $ are two roots. Since it is a cubic equation we need to have 3 roots. We know that irrational roots occur in conjugate pairs. Here $1 + \sqrt 2 $is an irrational root and hence the other root is $1 - \sqrt 2 $. Now we have the 3 roots of the cubic equation. We know that if a , b , c are the roots of a cubic equation then the cubic equation is given by
$ \Rightarrow {x^3} - \left( {a + b + c} \right){x^2} + \left( {ab + bc + ca} \right)x - abc = 0$
Hence here a = 3 and b = $1 + \sqrt 2 $and c =$1 - \sqrt 2 $
Hence our cubic equation is given by
$
   \Rightarrow {x^3} - \left( {3 + 1 + \sqrt 2 + 1 - \sqrt 2 } \right){x^2} + \left( {3\left( {1 + \sqrt 2 } \right) + \left( {1 + \sqrt 2 } \right)\left( {1 - \sqrt 2 } \right) + 3\left( {1 - \sqrt 2 } \right)} \right)x - 3\left( {1 + \sqrt 2 } \right)\left( {1 - \sqrt 2 } \right) = 0 \\
   \Rightarrow {x^3} - \left( {3 + 1 + 1} \right){x^2} + \left( {\left( {3 + 3\sqrt 2 } \right) + \left( {1 - 2} \right) + \left( {3 - 3\sqrt 2 } \right)} \right)x - 3\left( {1 - 2} \right) = 0 \\
   \Rightarrow {x^3} - 5{x^2} + \left( {3 + 3\sqrt 2 - 1 + 3 - 3\sqrt 2 } \right)x - 3\left( { - 1} \right) = 0 \\
   \Rightarrow {x^3} - 5{x^2} + 5x + 3 = 0 \\
 $
Hence this the cubic equations whose roots are 3 , $1 + \sqrt 2 $ and $1 - \sqrt 2 $

Therefore the correct option is d.

Note:
Whenever three roots a , b , c are given then the cubic equation can also be found by
$ \Rightarrow \left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right) = 0$
Hence here a = 3 and b = $1 + \sqrt 2 $ and c =$1 - \sqrt 2 $
Hence our cubic equation is given by
$ \Rightarrow \left( {x - 3} \right)\left( {x - \left( {1 + \sqrt 2 } \right)} \right)\left( {x - \left( {1 - \sqrt 2 } \right)} \right) = 0$
Multiplying this we get the required equation
\[
   \Rightarrow \left( {x - 3} \right)\left( {x - 1 - \sqrt 2 } \right)\left( {x - 1 + \sqrt 2 } \right) = 0 \\
   \Rightarrow \left( {x - 3} \right)\left( {{{\left( {x - 1} \right)}^2} - 2} \right) = 0 \\
   \Rightarrow \left( {x - 3} \right)\left( {{x^2} + 1 - 2x - 2} \right) = 0 \\
   \Rightarrow \left( {x - 3} \right)\left( {{x^2} - 2x - 1} \right) = 0 \\
   \Rightarrow {x^3} - 2{x^2} - x - 3{x^2} + 6x + 3 = 0 \\
   \Rightarrow {x^3} - 5{x^2} + 5x + 3 = 0 \\
 \]