Question

If one root of the quadratic equation is $\sqrt{3}+1$, then equation is
A. ${{x}^{2}}-2x-\sqrt{3}=0$
B. ${{x}^{2}}-2\sqrt{3}x+2=0$
C. ${{x}^{2}}-2x-2=0$
D. ${{x}^{2}}+2\sqrt{3}x+23=0$

Answer 1

Here we have given that an irrational/surd root of an equation. We know if $p+\sqrt{q}$ is an irrational/surd root of an equation, then $p-\sqrt{q}$ will be the other root of the equation. So, we can get the values of two roots of an equation. Now we will apply the relation between the roots of the equation and the coefficients of the equation i.e. if $\alpha,\beta$ are the roots of an equation $a{{x}^{2}}+bx+c=0$, then
$\alpha +\beta =\dfrac{-b}{a}$,$\alpha \beta =\dfrac{c}{a}$
From the above relation, we will get the equations. From those equations, we will verify the given options to find the correct option.

Complete step-by-step solution
Given that,
$\sqrt{3}+1$ is the root of an equation.
Let the equation is $a{{x}^{2}}+bx+c=0$
The given root $1+\sqrt{3}$ is an irrational root, hence the other root of the equation is $1-\sqrt{3}$. Hence
$\alpha =1+\sqrt{3}$, $\beta =1-\sqrt{3}$
Now the sum of the roots is
$\begin{aligned} & \alpha +\beta =1+\sqrt{3}+1-\sqrt{3} \\\ & =2 \end{aligned}$
But we know that $\alpha +\beta =\dfrac{-b}{a}$, hence
$\begin{aligned} & \dfrac{-b}{a}=2 \\\ & b=-2a....\left( \text{i} \right) \end{aligned}$
Now the product of the two roots is
$\alpha \beta =\left( 1+\sqrt{3} \right)\left( 1-\sqrt{3} \right)$
Using the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in the above equation, then
$\begin{aligned} & \alpha \beta ={{1}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \\\ & =1-3 \\\ & =-2 \end{aligned}$
But we know that $\alpha \beta =\dfrac{c}{a}$, hence
$\begin{aligned} & \dfrac{c}{a}=-2 \\\ & c=-2a....\left( \text{ii} \right) \end{aligned}$
From the $\left( \text{i} \right),\left( \text{ii} \right)$ we have $b=c$
So, checking the above condition in the given options
A. ${{x}^{2}}-2x-\sqrt{3}=0$ comparing with $a{{x}^{2}}+bx+c=0$, then
$a=1$, $b=-2$, $c=-\sqrt{3}$
Here $b\ne c$, so this is not the correct option.
B. ${{x}^{2}}-2\sqrt{3}x+2=0$ comparing with $a{{x}^{2}}+bx+c=0$, then
$a=1$, $b=-2\sqrt{3}$, $c=2$
Here $b\ne c$, so this is not the correct option.
C. ${{x}^{2}}-2x-2=0$ comparing with $a{{x}^{2}}+bx+c=0$, then
$a=1$, $b=-2$, $c=-2$
Here $b=c$, so this is the correct option.
D. ${{x}^{2}}+2\sqrt{3}x+23=0$ comparing with $a{{x}^{2}}+bx+c=0$, then
$a=1$, $b=2\sqrt{3}$, $c=23$
Here $b\ne c$, so this is not the correct option.
Hence the equation with $\sqrt{3}+1$as root is ${{x}^{2}}-2x-2=0$.

Note: Please note that the irrational roots should be in the form of $p+\sqrt{q}$. In the problem they mentioned $\sqrt{3}+1$, so we need to change it to $1+\sqrt{3}$ and take the other root as $1-\sqrt{3}$, do not take the other root as $\sqrt{3}-1$. We can also check the answer by substituting $x=1+\sqrt{3}$ or $x=1-\sqrt{3}$ in the answer.
The value of ${{x}^{2}}-2x-2$, when $x=1+\sqrt{3}$ is
$\begin{aligned} & {{x}^{2}}-2x-2={{\left( 1+\sqrt{3} \right)}^{2}}-2\left( 1+\sqrt{3} \right)-2 \\\ & ={{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+2.\left( 1 \right)\left( \sqrt{3} \right)-2-2\sqrt{3}-2 \\\ & =1+3+2\sqrt{3}-2\sqrt{3}-4 \\\ & =0 \end{aligned}$
Hence $x=1+\sqrt{3}$ is one root of ${{x}^{2}}-2x-2=0$.
The value of ${{x}^{2}}-2x-2$, when $x=1-\sqrt{3}$ is
$\begin{aligned} & {{x}^{2}}-2x-2={{\left( 1-\sqrt{3} \right)}^{2}}-2\left( 1-\sqrt{3} \right)-2 \\\ & ={{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2.\left( 1 \right)\left( \sqrt{3} \right)-2+2\sqrt{3}-2 \\\ & =1+3-2\sqrt{3}+2\sqrt{3}-4 \\\ & =0 \end{aligned}$
Hence $x=1-\sqrt{3}$ is one root of ${{x}^{2}}-2x-2=0$.