Question

If $\sqrt{2}+1$is a root of a quadratic equation with rational coefficients, then write its other root.

Answer 1

Hint: First, before proceeding for this, we must know that the formula for the sum of roots of the quadratic equation of form $a{{x}^{2}}+bx+c=0$ where $\alpha $and $\beta $are roots is given by $\alpha +\beta =\dfrac{-b}{a}$. Then, by substituting the value of one root let us suppose be $\alpha =1+\sqrt{2}$be given root and we need to find $\beta $in the above expression, we have left hand side with irrational number $\sqrt{2}$which proves that second root is something which cancels this value. Then, by using the formula for product of roots for the same quadratic equation $a{{x}^{2}}+bx+c=0$ where $\alpha $and $\beta $are roots is given by $\alpha \times \beta =\dfrac{c}{a}$, we get the value of other root.

Complete step by step answer:
In this question, we are supposed to find the value of the other root of the quadratic equation when one of its roots is $\sqrt{2}+1$.
So, before proceeding for this, we must know that the formula for the sum of roots of the quadratic equation of form $a{{x}^{2}}+bx+c=0$ where $\alpha $and $\beta $are roots is given by:
$\alpha +\beta =\dfrac{-b}{a}$
Now, by substituting the value of one root let us suppose be $\alpha =1+\sqrt{2}$be given root and we need to find $\beta $in the above expression as:
$1+\sqrt{2}+\beta =\dfrac{-b}{a}$
Now, we can see clearly that right hand side is purely rational number without any irrational part.
However, we have left-hand side with an irrational number $\sqrt{2}$which proves that the second root is something that cancels this value.
So, we get the value of $\beta $ as of form $A-\sqrt{2}$where A is any constant value that does not gives a conclusive answer.
Then, by using the formula for product of roots for the same quadratic equation $a{{x}^{2}}+bx+c=0$ where $\alpha $and $\beta $are roots is given by:
$\alpha \times \beta =\dfrac{c}{a}$
Now, by substituting the value of $\alpha $and $\beta $in above expression, we get:
$\begin{align}
  & \left( 1+\sqrt{2} \right)\times \left( A-\sqrt{2} \right)=\dfrac{c}{a} \\
 & \Rightarrow A-\sqrt{2}+\sqrt{2}A-2=\dfrac{c}{a} \\
 & \Rightarrow \left( A-2 \right)+\left( A-1 \right)\sqrt{2}=\dfrac{c}{a} \\
\end{align}$
So, we can see clearly that right hand side only has rational part which means the irrational part on the left hand side should be zero.
So, by equation the irrational part of left hand side with zero, we get:
$\begin{align}
  & \sqrt{2}\left( A-1 \right)=0 \\
 & \Rightarrow A-1=0 \\
 & \Rightarrow A=1 \\
\end{align}$
So, we get the value of A as 1 and which in turn gives the value of other root $\beta $as:
$\begin{align}
  & \beta =A-\sqrt{2} \\
 & \Rightarrow \beta =1-\sqrt{2} \\
\end{align}$
Hence, we get the value of other root as $1-\sqrt{2}$.

Note:
 Now, to solve these type of the questions we need to know some of the shortcuts also so that we can get the answer within seconds that most of the time if one root of the quadratic equation is of form $a+\sqrt{2}$, then the other root of the equation is mostly as $a-\sqrt{2}$.