Question

If one root is $3+\sqrt{5}$, then the quadratic equation will be:

• A. $x^2+6x-4=0$
• B. $x^2+6x+4=0$
• C. $x^2-6x+4=0$
• D. $x^2-6x-4=0$

Answer 1

The correct option is C

$x^2-6x+4=0$

Step 1: Given data.

It is given that one root of the quadratic equation is $3+\sqrt{5}$.

The objective is to find the quadratic equation.

Step 2: Apply the quadratic formula to find the quadratic equation.

Let $\alpha$ and $\beta$ be the two roots of the quadratic equation $a{x}^2+bx+c=0$.

Since $a{x}^2+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

Let $\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$.

Now, compare the given root $\frac{3+\sqrt{5}}{1}$ with the expression $\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$ to get:

$\begin{array}{rcl}-b& =& 3\\ & \Rightarrow & b=-3\end{array}$

Also, $2a=1$

$\Rightarrow a=\frac{1}{2}$

Similarly, we have $b^2-4ac=5$

$\begin{array}{rcl}& \Rightarrow & {\left(-3\right)}^2-4\left(\frac{1}{2}\right)c=5\\ & \Rightarrow & 9-2c=5\\ & \Rightarrow & 2c=4\\ & \Rightarrow & c=2\end{array}$

Step 3: Final answer.

Substituting the values of $a,b$ and $c$ in the quadratic equation $a{x}^2+bx+c=0$ as follows:

$\begin{array}{rcl}& \Rightarrow & \left(\frac{1}{2}\right)x^2+\left(-3\right)x+2=0\\ & \Rightarrow & x^2-6x+4=0\end{array}$

Therefore, $x^2-6x+4=0$ is the required quadratic equation.

Hence, the correct option is C.