Question

[Maths]

$\large \underbrace{ \overbrace{ \over \underline{\texttt{Question:-}}}}$
Find the nature of the following quadratic equations:

$\dag \: \: \green{ \tt {x}^{2} - 2 \sqrt{3} - 1 = 0 }$
If real roots exists, find them.​

Answer 1

Answer:

• Roots are real and distinct.
• Real root which exists are √3 ± 2.

Step-by-step explanation:

Given

• x² - 2√3x - 1 = 0

To find

• Nature of the quadratic equation.
• If real roots exists, find them.

Solution

Let's find out the discriminant.

✮The discriminant of a quadratic equation tells us whether there are two solutions, one solution, or no solutions.

⇾ D = b² - 4ac

⇾ x² - 2√3x - 1 = 0

Comparing with quadratic equation :

• ax² + bx + c = 0

We get :

• a = 1
• b = -2√3
• c = -1

With the discriminant :

• b² - 4ac

We get :

• b = -2√3
• a = 1
• c = -1

Substituting we get :

• (-2√3)² - 4(1)(-1)
• (-2√3 × -2√3) - 4(-1)
• (-2 × -2) (√3 × √3) + 4
• (4) (3) + 4
• 12 + 4
• 16

Hence, the discriminant is 16.

And it is > 0 , hence the roots are real and distinct.

Real roots can be found using Sridacharya formula,

• x = -b ± √b² - 4ac / 2a

Substituting we get :

• x = 2√3 ± √16 / 2
• 2√3 ± 4 / 2
• √3 ± 2

Hence, the real root which exists are √3 ± 2.

Know these

• If D > 0 - roots are real and distinct.
• If D < 0 - roots are imaginary.
• If D = 0 - roots are equal.

Answer 2

Correct question

$x^2-2\sqrt{3} x-1=0$

Required method

If we have a quadratic equation, we can use the discriminant to find the nature of the roots. Since discriminant is placed inside the square root, the nature of the roots depends on it. Here, $D$ is the discriminant.

Solution

We have a quadratic equation in the form of $ax^2+bx+c=0$.

Comparing to the equation,

• $a=1$
• $b=-2\sqrt{3}$
• $c=-1$

Now, let's find the discriminant.

Comparing to $D=b^2-4ac$,

$D=(-2\sqrt{3} )^2-4\times1\times(-1)$

   $=12+4$

   $=16>0$

Since $D>0$, the roots are real and different.

Comparing to the quadratic formula $x=\dfrac{-b\pm\sqrt{D} }{2a}$,

$x=\dfrac{2\sqrt{3} \pm\sqrt{16} }{2}$

  $=\dfrac{2\sqrt{3} \pm4}{2}$

  $=\sqrt{3} \pm2$

So, real roots exist and both are $\sqrt{3} \pm2$.

More information

Consider a quadratic equation $ax^2+bx+c=0$.

Let's also consider $D=b^2-4ac$.

If $ac<0$, the discriminant will always be positive, so it has real roots.