MCQ
Which of the following equations has no real roots?
- ✓$x^2-4 x+3 \sqrt{2}=0$
- B$x^2+4 x-3 \sqrt{2}=0$
- C$x^2-4 x-3 \sqrt{2}=0$
- D$3 x^2+4 \sqrt{3} x+4=0$
✓
Answer
Correct option: A.
$x^2-4 x+3 \sqrt{2}=0$
$x^2-4 x+3 \sqrt{2}=0$
Explanation:
(A) The given equation is $x^2-4 x+3 \sqrt{2}=0$
On comparing with $ax^2 + bx + c = 0,$ we get
$a = 1, b = – 4$ and $c =3 \sqrt{2}$
The discriminant of $x^2-4 x+3 \sqrt{2}=0$ is
$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$
$b^2 – 4ac < 0$
$(B)$ The given equation is$x^2+4 x-3 \sqrt{2}=0$
On comparing the equation with $ax^2 + bx + c = 0$, we get
$a = 1, b = 4$ and $c =-3 \sqrt{2}$
Then, $D = b^2 – 4ac$
$=(-4)^2-4(1)(-3 \sqrt{2}) $
$ =16+12 \sqrt{2}>0$
Hence, the equation has real roots.
(C) Given equation is $x^2-4 x-3 \sqrt{2}=0$
On comparing the equation with $ax^2 + bx + c = 0,$ we get
$a = 1, b = – 4$ and $c =-3 \sqrt{2}$
Then, $D = b^2 – 4ac$
$=(-4)^2-4(1)(-3 \sqrt{2}) $
$ =16+12 \sqrt{2}>0$
Hence, the equation has real roots.
$(D)$ Given equation is $3 x^2+4 \sqrt{3} x+4=0$
On comparing the equation with $ax^2 + bx + c = 0,$ we get
$a = 3, b =4 \sqrt{3}$ and $c=4$
Then, $D = b^2 – 4ac$
$=(4 \sqrt{3})^2-4(3)(4)$
$=48-48 $
$=0$
Hence, the equation has real roots.
Hence, $x^2-4 x+3 \sqrt{2}=0$ has no real roots.
Explanation:
(A) The given equation is $x^2-4 x+3 \sqrt{2}=0$
On comparing with $ax^2 + bx + c = 0,$ we get
$a = 1, b = – 4$ and $c =3 \sqrt{2}$
The discriminant of $x^2-4 x+3 \sqrt{2}=0$ is
$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$
$b^2 – 4ac < 0$
$(B)$ The given equation is$x^2+4 x-3 \sqrt{2}=0$
On comparing the equation with $ax^2 + bx + c = 0$, we get
$a = 1, b = 4$ and $c =-3 \sqrt{2}$
Then, $D = b^2 – 4ac$
$=(-4)^2-4(1)(-3 \sqrt{2}) $
$ =16+12 \sqrt{2}>0$
Hence, the equation has real roots.
(C) Given equation is $x^2-4 x-3 \sqrt{2}=0$
On comparing the equation with $ax^2 + bx + c = 0,$ we get
$a = 1, b = – 4$ and $c =-3 \sqrt{2}$
Then, $D = b^2 – 4ac$
$=(-4)^2-4(1)(-3 \sqrt{2}) $
$ =16+12 \sqrt{2}>0$
Hence, the equation has real roots.
$(D)$ Given equation is $3 x^2+4 \sqrt{3} x+4=0$
On comparing the equation with $ax^2 + bx + c = 0,$ we get
$a = 3, b =4 \sqrt{3}$ and $c=4$
Then, $D = b^2 – 4ac$
$=(4 \sqrt{3})^2-4(3)(4)$
$=48-48 $
$=0$
Hence, the equation has real roots.
Hence, $x^2-4 x+3 \sqrt{2}=0$ has no real roots.
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