Question

Which of the following equations has no real roots?

(A) $x^2-4x+3\sqrt{2}=0$

(B) $x^2+4x-3\sqrt{2}=0$

(C) $x^2-4x-3\sqrt{2}=0$

(D) $3x^2+4\sqrt{3}x+4=0$

• A. ( A )
• B. ( B )
• C. ( C )
• D. ( D )

Answer 1

The correct option is A

( A )

Explanation for the correct option:

Option (A): Consider $x^2-4x+3\sqrt{2}=0$.

The quadratic equation $a{x}^2+bx+c=0$ has no real roots if $b^2-4ac<0$.

$\begin{array}{rcl}{4}^2-4\left(1\right)\left(3\sqrt{2}\right)& =& 0\\ & \Rightarrow & 16-16.9705=0\\ & \Rightarrow & -0.9705<0\end{array}$

Hence, option (A) is correct.

Explanation for the incorrect options:

Option (B): Consider $x^2+4x-3\sqrt{2}=0$.

$$\begin{gathered} \begin{array}{rcl}{4}^2-4\left(1\right)\left(-3\sqrt{2}\right)& =& 0\\ & \Rightarrow & 16+16.9705=0\end{array} \\ \Rightarrow \begin{array}{rcl}& & 32\end{array}\begin{array}{rcl}& & .\end{array}\begin{array}{rcl}& & 9705\end{array}\begin{array}{rcl}& >& \end{array}\begin{array}{rcl}& & 0\end{array} \end{gathered}$$

Therefore, option (B) is incorrect.

Option (C): Consider $x^2-4x-3\sqrt{2}=0$.

$$\begin{gathered} \begin{array}{rcl}{\left(-4\right)}^2-4\left(1\right)\left(-3\sqrt{2}\right)& =& 0\\ & \Rightarrow & 16+16.9705=0\end{array} \\ \Rightarrow 32.9705>0 \end{gathered}$$

Therefore, option (C) is incorrect.

Option (D): Consider $3x^2+4\sqrt{3}x+4=0$.

Substitute $x=2$ into the given equation to find whether the equation has $2$ as a root.

$$\begin{gathered} {\left(4\sqrt{3}\right)}^2-4\left(3\right)\left(4\right)=0 \\ \Rightarrow 48-48=0 \\ \Rightarrow 0=0 \end{gathered}$$

Therefore, option (D) is incorrect.

Thus, options (B), (C), and (D) are incorrect.

Hence, option (A) is correct.