Question

Question 10
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) $3x^2-4\sqrt{3}x+4=0$

Answer 1

a) The given equation is $x^2-4x+3\sqrt{2}=0$
On comparing with $a{x}^2+bx+c=0$, we get
A = 1, b = -4 and c = $3\sqrt{2}$
The discriminant of $x^2–4x+3\sqrt{2}$ = 0 is
$D=b^2-4ac$
$=(-4)2-4\left(1\right)\left(3\sqrt{2}\right)=16-12\sqrt{2}=16-12\times \left(1.41\right)$
$=16-16.92=-0.92$
$b^2-4ac<0$

b) The given equation is $x^2+4x-3\sqrt{2}=0$
On comparing with $a{x}^2+bx+c=0$, we get
A = 1, b = 4 and c = $-3\sqrt{2}$
The discriminant of $x^2+4x-3\sqrt{2}$ = 0 is
$D=b^2-4ac$
$=\left(4\right)2-4\left(1\right)(-3\sqrt{2})=16+12\sqrt{2}=16+12\times \left(1.41\right)$
$=16+16.92=32.92$
$b^2-4ac>0$

c) The given equation is $x^2-4x-3\sqrt{2}=0$
On comparing with $a{x}^2+bx+c=0$, we get
A = 1, b = -4 and c = $-3\sqrt{2}$
The discriminant of $x^2-4x-3\sqrt{2}$ = 0 is
$D=b^2-4ac$
$=(-4)2-4\left(1\right)(-3\sqrt{2})=16+12\sqrt{2}=16+12\times \left(1.41\right)$
$=16+16.92=32.92$
$b^2-4ac>0$

d) The given equation is $3x^2$ – $4\sqrt{3}x+1=0$
on comparing with $a{x}^2+bx+c=0$, we get
A = 3, B = $4\sqrt{3}$ and C = 4
The discriminant of $3x^2-4\sqrt{3}x+4=0$
$D=b^2-4ac$
$=(4\sqrt{3}{)}^2-4(4\left)\right(3)=48-48=0$

Therefore, only equation given in option (A) has no real roots.