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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.