Problem 52 - Contemporary Precalculus

Question

Solve the equation by any method. $$x^{2}+\sqrt{2} x-3=0$$

Step-by-Step Solution

Answer

Answer: The solutions for the given quadratic equation are $$x=\frac{-\sqrt{2}+\sqrt{14}}{2}$$ and $$x=\frac{-\sqrt{2}-\sqrt{14}}{2}$$.

Step 1: Identify the coefficients a, b, and c

The given quadratic equation is $$x^{2}+\sqrt{2} x-3=0$$. We can see that the coefficents are: $$a=1$$, $$b=\sqrt{2}$$, $$c=-3$$.

Step 2: Plug the coefficients into the Quadratic Formula

The Quadratic Formula is $$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$. We will now substitute a, b, and c with their respective values: $$x=\frac{-\sqrt{2}\pm\sqrt{(\sqrt{2})^{2}-4(1)(-3)}}{2(1)}$$

Step 3: Simplify the expression

Now, let's simplify the expression to find the values of x: $$x=\frac{-\sqrt{2}\pm\sqrt{2+12}}{2}$$ $$x=\frac{-\sqrt{2}\pm\sqrt{14}}{2}$$ Now, we have 2 possible solutions for x:

Step 4: Determine the two possible solutions

Using the simplified quadratic formula, we calculate the two possible solutions for x: 1) $$x=\frac{-\sqrt{2}+\sqrt{14}}{2}$$ 2) $$x=\frac{-\sqrt{2}-\sqrt{14}}{2}$$ So, the solution to the given equation $$x^{2}+\sqrt{2} x-3=0$$ is: 1) $$x=\frac{-\sqrt{2}+\sqrt{14}}{2}$$ 2) $$x=\frac{-\sqrt{2}-\sqrt{14}}{2}$$

Key Concepts

Quadratic Formula

The Quadratic Formula is a powerful tool used to find solutions of quadratic equations. Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. This is a universal formula because it works for any quadratic equation.
Transforming a quadratic equation into numbers that we can easily manipulate requires the use of the Quadratic Formula which is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this formula:

• \( b \) represents the coefficient of the \( x \) term.
• \( a \) is the coefficient before \( x^2 \).
• \( c \) is the constant term.

The term inside the square root, \( b^2 - 4ac \), is known as the discriminant. It helps in determining the nature and number of roots. If the discriminant is positive, the equation has two distinct real roots. If it's zero, there is one real double root. If it's negative, there are no real roots, only complex ones.

Coefficients

In any quadratic equation of the form \( ax^2 + bx + c = 0 \), the numbers \( a \), \( b \), and \( c \) are known as coefficients. They tell us a lot about the equation.

• \( a \) is the coefficient of \( x^2 \), and it determines the shape and direction of the parabola that represents the equation. If \( a \) is positive, the parabola opens upwards, and if \( a \) is negative, it opens downwards.
• \( b \) is the coefficient of \( x \), and it affects the symmetry and position of the parabola along the x-axis.
• \( c \) is the constant term, and it represents the y-intercept, which is where the parabola crosses the y-axis.

Each of these coefficients plays a crucial role in solving the quadratic equation using the Quadratic Formula, as they are the values you substitute into the formula to find solutions.

Solutions of Equations

Finding solutions of a quadratic equation means calculating the values of \( x \) that make the equation true. These are also known as roots or zeros of the equation.The Quadratic Formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used to find these solutions. For the given equation \( x^2 + \sqrt{2} x - 3 = 0 \):

• We substitute \( a = 1 \), \( b = \sqrt{2} \), and \( c = -3 \) into the formula.
• Simplifying the discriminant, \( b^2 - 4ac = (\sqrt{2})^2 - 4(1)(-3) = 2 + 12 = 14 \), gives us \( \pm \sqrt{14} \).
• This results in two possible solutions for \( x \):
• \( x = \frac{-\sqrt{2} + \sqrt{14}}{2} \)
• \( x = \frac{-\sqrt{2} - \sqrt{14}}{2} \)

These solutions are the values of \( x \) where the given quadratic equation equals zero. The solutions can be verified by substituting them back into the original equation to ensure they satisfy the equation.