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# Choose the correct answer from the given four options in the following questions:

Which of the following equations has no real roots?

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

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

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

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

To do:

We have to find the correct answer.

Solution:

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

Comparing with $a x^{2}+b x+c=0$, we get,

$a=1, b=-4$ and $c=3 \sqrt{2}$

$D=b^{2}-4 a c$

$=(-4)^{2}-4(1)(3 \sqrt{2})$

$=16-12 \sqrt{2}$

$=16-12 \times(1.41)$

$=16-16.92$

$=-0.92<0$

Hence, the given equation has no real roots.

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

Comparing the equation with $a x^{2}+b x+c=0$, we get,

$a=1, b=4$ and $c=-3 \sqrt{2}$

$D=b^{2}-4 a c$

$=(-4)^{2}-4(1)(-3 \sqrt{2})$

$=16+12 \sqrt{2}>0$

Hence, the equation has real roots.

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

Comparing the equation with $a x^{2}+b x+c=0$, we get,

$a =1, b=-4$ and $c=-3 \sqrt{2}$

$D=b^{2}-4 a c$

$=(-4)^{2}-4(1)(-3 \sqrt{2})$

$=16+12 \sqrt{2}>0$

Hence, the equation has real roots.

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

Comparing the equation with $a x^{2}+b x+c=0$, we get,

$a =3, b=4\sqrt3$ and $c=4$

$D=b^{2}-4 a c$

$=(4\sqrt3)^{2}-4(3)(4)$

$=48-48=0$

Hence, the equation has real roots.

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