Given: A quadratic equation $2x^2 - 6x + 3 = 0$.

To Find: The nature of the roots of the given quadratic equation and, if real roots exist, find their values.

Step 1: Identify the coefficients of the quadratic equation.

A standard quadratic equation is represented as $ax^2 + bx + c = 0$. Comparing the given equation $2x^2 - 6x + 3 = 0$ with the standard form:

• $a = 2$
• $b = -6$
• $c = 3$

Step 2: Determine the nature of the roots using the Discriminant ($D$).

The discriminant of a quadratic equation is given by the formula $D = b^2 - 4ac$.

Substituting the values of $a$, $b$, and $c$:

$D = (-6)^2 - 4(2)(3)$

$D = 36 - 24$

$D = 12$

[Since $D > 0$, the quadratic equation has two distinct real roots.]

Step 3: Apply the Quadratic Formula to find the roots.

The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.

Substituting the known values into the formula:

$x = \frac{-(-6) \pm \sqrt{12}}{2(2)}$

$x = \frac{6 \pm \sqrt{4 \times 3}}{4}$

$x = \frac{6 \pm 2\sqrt{3}}{4}$

Step 4: Simplify the expression.

Factor out the common term $2$ from the numerator:

$x = \frac{2(3 \pm \sqrt{3})}{4}$

$x = \frac{3 \pm \sqrt{3}}{2}$

Therefore, the two roots are:

$x_1 = \frac{3 + \sqrt{3}}{2}$

$x_2 = \frac{3 - \sqrt{3}}{2}$

Final Answer: The roots are real and distinct. The roots of the equation are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.

Given: The equation $(x^3 - 4x^2 - x + 1) = (x - 2)^3$.

To Find: Determine whether the given equation is a quadratic equation.

Step 1: Understanding the definition of a Quadratic Equation
A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$. The highest power (degree) of the variable $x$ must be exactly 2.

Step 2: Expanding the right-hand side (RHS)
The RHS is $(x - 2)^3$. We use the algebraic identity for the cube of a binomial:
$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
Here, $a = x$ and $b = 2$.
$(x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - (2)^3$
$(x - 2)^3 = x^3 - 6x^2 + 3(x)(4) - 8$
$(x - 2)^3 = x^3 - 6x^2 + 12x - 8$ [Applying the identity $(a-b)^3$]

Step 3: Equating LHS and RHS
Substitute the expanded RHS back into the original equation:
$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$

Step 4: Simplifying the equation
To simplify, we bring all terms to the left-hand side (LHS):
$(x^3 - 4x^2 - x + 1) - (x^3 - 6x^2 + 12x - 8) = 0$
$x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$ [Distributing the negative sign]

Step 5: Combining like terms
Group the terms by their powers of $x$:
$(x^3 - x^3) + (-4x^2 + 6x^2) + (-x - 12x) + (1 + 8) = 0$
$0x^3 + 2x^2 - 13x + 9 = 0$
$2x^2 - 13x + 9 = 0$

Step 6: Conclusion
The resulting equation is $2x^2 - 13x + 9 = 0$. This equation is in the form $ax^2 + bx + c = 0$, where $a = 2$, $b = -13$, and $c = 9$. Since the highest degree of the variable $x$ is 2 and $a \neq 0$, the equation satisfies the definition of a quadratic equation.

Final Answer: Yes, the given equation is a quadratic equation.

Given: A quadratic equation $3x^2 - 4\sqrt{3}x + 4 = 0$.

To Find: The nature of the roots and the roots themselves if they exist.

Step 1: Identify the coefficients of the quadratic equation.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Comparing the given equation $3x^2 - 4\sqrt{3}x + 4 = 0$ with the standard form, we identify:
$a = 3$
$b = -4\sqrt{3}$
$c = 4$

Step 2: Determine the nature of the roots using the Discriminant ($D$).
The discriminant is given by the formula $D = b^2 - 4ac$.
Substituting the values identified in Step 1:
$D = (-4\sqrt{3})^2 - 4(3)(4)$
$D = (16 \times 3) - 48$
$D = 48 - 48$
$D = 0$
[Since $D = 0$, the quadratic equation has two equal real roots.]

Step 3: Calculate the roots using the Quadratic Formula.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.
Since $D = 0$, the formula simplifies to $x = \frac{-b}{2a}$.
Substituting the values:
$x = \frac{-(-4\sqrt{3})}{2(3)}$
$x = \frac{4\sqrt{3}}{6}$

Step 4: Simplify the expression.
$x = \frac{4\sqrt{3}}{6}$
Dividing both the numerator and the denominator by their greatest common divisor, which is $2$:
$x = \frac{2\sqrt{3}}{3}$
Since the roots are equal, both roots are $\frac{2\sqrt{3}}{3}$.
[Note: $\frac{2\sqrt{3}}{3}$ can also be written as $\frac{2}{\sqrt{3}}$ by rationalizing the denominator.]

Final Answer: The roots are real and equal, and the roots are $\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$.