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.\[{x^2} + 5x + 3\sqrt 2 = 0\]
D.\[3{x^2} + 4\sqrt 3 x + 4 = 0\]

Answer 1

Hint: Here we will use the concept of discriminant of a quadratic equation. We will find the discriminant of the quadratic equation in each given option. We will find the nature of the roots using the properties of the discriminant.

Formula Used:
If a quadratic equation is of the form of \[a{x^2} + bx + c = 0\] , then we will use the discriminant formula, \[D = {b^2} - 4ac\].

Complete step-by-step answer:
We have to find which of the given equations has no real roots.
As we know, if a quadratic equation has no real roots then, the discriminant is less than zero.
\[D < 0\]
\[ \Rightarrow {b^2} - 4ac < 0\]
Now we will calculate the discriminant of each equation.
\[{x^2} + 4x + 3\sqrt 2 = 0\]
Comparing this equation with \[a{x^2} + bx + c = 0\], we get
\[a = 1\], \[b = 4\] and \[c = 3\sqrt 2 \]
Now, substituting \[a = 1\], \[b = 4\] and \[c = 3\sqrt 2 \] in the formula \[D = {b^2} - 4ac\], we get
\[ \Rightarrow D = {\left( 4 \right)^2} - 4\left( 1 \right)\left( {3\sqrt 2 } \right)\]
Solving further, we get
\[ \Rightarrow D = 16 - 12\sqrt 2 \]
Substituting the value of \[\sqrt 2 = 1.41\] in the above equation, we get
\[ \Rightarrow D = 16 - 12\left( {1.41} \right)\]
Simplifying the terms, we get
\[ \Rightarrow D = 16 - 16.92 = - 0.92\]
We can see that \[D < 0\].
Hence, the equation \[{x^2} + 4x + 3\sqrt 2 = 0\] has no real roots.
Therefore, option A is the correct option.
But, if we check the other equations:
\[{x^2} + 4x - 3\sqrt 2 = 0\]
Comparing this equation with \[a{x^2} + bx + c = 0\], we get
Now, substituting \[a = 1{\rm{, }}b = 4\] and \[{\rm{ }}c = - 3\sqrt 2 \] in the formula \[D = {b^2} - 4ac\], we get
 \[D = {\left( 4 \right)^2} - 4\left( 1 \right)\left( { - 3\sqrt 2 } \right)\]
Solving further, we get,
\[ \Rightarrow D = 16 + 12\sqrt 2 \]
Here, As, \[D > 0\] , hence, real roots exist.
\[{x^2} + 5x + 3\sqrt 2 = 0\]
Comparing this equation with \[a{x^2} + bx + c = 0\], we get
\[a = 1{\rm{, }}b = 5\] and \[c = 3\sqrt 2 \]
Now, substituting \[a = 1{\rm{, }}b = 5\] and \[c = 3\sqrt 2 \] in the formula \[D = {b^2} - 4ac\], we get
\[D = {\left( 5 \right)^2} - 4\left( 1 \right)\left( {3\sqrt 2 } \right)\]
Solving further, we get,
\[ \Rightarrow D = 25 - 12\sqrt 2 \]
Substituting the value of \[\sqrt 2 = 1.41\] in the above equation, we get
\[D = 25 - 12\left( {1.41} \right)\]
$\Rightarrow$ \[D = 25 - 16.92 = 8.08\]
As, \[D > 0\], hence, real roots exist.
\[3{x^2} + 4\sqrt 3 x + 4 = 0\]
Comparing this equation with \[a{x^2} + bx + c = 0\], we get
\[a = 3{\rm{, }}b = 4\sqrt 3 \] and \[c = 4\]
Now, substituting \[a = 3{\rm{, }}b = 4\sqrt 3 \] and \[c = 4\] in the formula \[D = {b^2} - 4ac\], we get
\[D = {\left( {4\sqrt 3 } \right)^2} - 4\left( 3 \right)\left( 4 \right)\]
Solving further, we get,
$\Rightarrow$ \[D = 48 - 48 = 0\]
As, \[D = 0\], hence, equal roots exist.
Therefore, option A is the required answer as it is the only equation which has no real roots.

Note: Here we are provided with the quadratic equation. Quadratic equation is an equation which has a highest degree of 2. While solving this question, we should know the formula and properties of Discriminant and we should keep in mind that:
We get two distinct roots i.e. positive and different when \[D > 0\].
We get to equal roots i.e. positive and same when \[D = 0\].
We get no real roots i.e. negative roots when \[D < 0\].