Problem 8 Solve using the quadratic formul... [FREE SOLUTION]

Chapter 8: Problem 8

Solve using the quadratic formula. a. \(x^{2}-3 x=12\) e. \(\frac{1}{x-2}=\frac{x+1}{x-1}\) b. \(3 x^{2}=4 x+2\) f. \(\frac{x^{2}}{3}+\frac{x}{2}-\frac{1}{6}=0\) c. \(3\left(x^{2}+1\right)=x+2\) g. \(\frac{1}{x^{2}}-\frac{3}{x}=\frac{1}{6}\) d. \((3 x-1)(x+2)=4\)

Short Answer

Expert verified

a) 5, -2 b) 2, -2/3 c) No real solutions d) 1, -11/3 e) 2.41, -0.41 f) 0.4, -1.9 g) 0.08, -18.08

Step by step solution

Move all terms to one side

Rewrite each equation in the standard quadratic form: \( ax^2 + bx + c = 0 \).

a. Rewrite the Equation

Start with \( x^2 - 3x = 12 \).Subtract 12 from both sides to get \( x^2 - 3x - 12 = 0 \).

b. Rewrite the Equation

Start with \( 3x^2 = 4x + 2 \).Move all terms to one side: \( 3x^2 - 4x - 2 = 0 \).

c. Rewrite the Equation

Start with \( 3(x^2 + 1) = x + 2 \).Expand and move terms: \( 3x^2 + 3 - x - 2 = 0 \) which simplifies to \( 3x^2 - x + 1 = 0 \).

d. Rewrite the Equation

Start with \( (3x-1)(x+2) = 4 \).Expand and move terms: \( 3x^2 + 6x - x - 2 - 4 = 0 \) which simplifies to \( 3x^2 + 5x - 6 = 0 \).

e. Rewrite the Equation

Start with \( \frac{1}{x-2} = \frac{x+1}{x-1} \).Cross-multiply: \( (x-1) = (x+1)(x-2) \).Expand and move terms: \( x - 1 = x^2 - x - 2 \) which simplifies to \( x^2 - 2x - 1 = 0 \).

f. Rewrite the Equation

Start with \( \frac{x^2}{3} + \frac{x}{2} - \frac{1}{6} = 0 \).Multiply by 6 to clear the fractions: \( 2x^2 + 3x - 1 = 0 \).

g. Rewrite the Equation

Start with \( \frac{1}{x^2} - \frac{3}{x} = \frac{1}{6} \).Get common denominator \( 6x^2 \): \( 6 -18x = x^2 \) simplifies to \( x^2 + 18x - 6 = 0 \).

Identify coefficients

For each equation, identify coefficients \( a, b, \) and \( c \).

a. Identify coefficients

Here, \( a = 1, b = -3, c = -12 \).

b. Identify coefficients

Here, \( a = 3, b = -4, c = -2 \).

c. Identify coefficients

Here, \( a = 3, b = -1, c = 1 \).

d. Identify coefficients

Here, \( a = 3, b = 5, c = -6 \).

e. Identify coefficients

Here, \( a = 1, b = -2, c = -1 \).

f. Identify coefficients

Here, \( a = 2, b = 3, c = -1 \).

g. Identify coefficients

Here, \( a = 1, b = 18, c = -6 \).

Apply the Quadratic Formula

Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

a. Apply Quadratic Formula

Substitute into the formula: \( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-12)}}{2(1)} = \frac{3 \pm \sqrt{9 + 48}}{2} = \frac{3 \pm 7}{2} \). Thus, \( x = 5 \) or \( x = -2 \).

b. Apply Quadratic Formula

Substitute into formula: \( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-2)}}{2(3)} = \frac{4 \pm \sqrt{16 + 24}}{6} = \frac{4 \pm 8}{6} \). Thus, \( x = 2 \) or \( x = -\frac{2}{3} \).

c. Apply Quadratic Formula

Substitute into the formula: \( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(1)}}{2(3)} = \frac{1 \pm \sqrt{1 - 12}}}{6} = \frac{1 \pm \sqrt{-11}}{6} \). Thus, \( x \) has no real solutions.

d. Apply Quadratic Formula

Substitute into formula: \( x = \frac{-5 \pm \sqrt{5^2 - 4(3)(-6)}}{2(3)} = \frac{-5 \pm \sqrt{25 + 72}}{6} = \frac{-5 \pm 11}{6} \). Thus, \( x = 1 \) or \( x = -\frac{11}{3} \).

e. Apply Quadratic Formula

Substitute into formula: \( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2.83}{2} \). Thus, \( x = 2.41 \) or \( x = -0.41 \).

f. Apply Quadratic Formula

Substitute into formula: \( x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm 4.58}{4} \). Thus, \( x = 0.4 \) or \( x = -1.9 \).

g. Apply Quadratic Formula

Substitute into formula: \( x = \frac{-18 \pm \sqrt{18^2 - 4(1)(6)}}{2(1)} = \frac{-18 \pm \sqrt{324 + 24}}{2} = \frac{-18 \pm 18.17}{2} \). Thus, \( x = 0.08 \) or \( x = -18.08 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving quadratic equations

A quadratic equation generally has the form \[ax^2 + bx + c = 0\].Solving these equations involves a specific process. Let's break it down:

Step 1: Rewrite the equation in standard form
First, make sure all terms are on one side of the equation in the form: \[ax^2 + bx + c = 0\]. This might involve moving terms across the equal sign and combining like terms.
For example, if you start with \[x^2 - 3x = 12\], you need to subtract 12 from both sides to rewrite it as \[x^2 - 3x - 12 = 0\].
Each exercise will involve this reformatting step. Make sure you are comfortable with moving terms to achieve this standard form.

Example (a): Convert \[3x^2 = 4x + 2\] to \[3x^2 - 4x - 2 = 0\] by subtracting \[4x\] and \[2\] from both sides.
Example (b): Expand and simplify \[3(x^2 + 1) = x + 2\] to get \[3x^2 + 3 - x - 2 = 0\], further simplified to \[3x^2 - x + 1 = 0\].

identifying coefficients

Once you have your equation in the standard form \[ax^2 + bx + c = 0\], you can easily identify the coefficients:

\forall\text: \[a\] is the coefficient of \[x^2\]
\forall\text: \[b\] is the coefficient of \[x\]
\forall\text: \[c\] is the constant term.

To illustrate: In the equation \[x^2 - 3x - 12 = 0\], the coefficients are \[a = 1\], \[b = -3\], and \[c = -12\].
Identifying these coefficients is crucial because you will need them to apply the quadratic formula.

Example (c): For \[3x^2 - 4x - 2 = 0\], the coefficients are \[a = 3\], \[b = -4\], \[c = -2\].
Example (d): In \[3x^2 + 5x - 6 = 0\], \[a = 3\], \[b = 5\], and \[c = -6\].

applying the quadratic formula

The quadratic formula is a powerful tool to solve any quadratic equation. Once you have identified the coefficients \[a, b,\text{ and } c\], plug them into the formula:
\[\begin{equation} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{equation}\]

First, calculate the discriminant, which is the part under the square root: \[b^2 - 4ac\]
Then, handle the rest of the formula to find the roots.

Steps:

Calculate \(-b\)|
Compute \(\root{b^2 - 4ac}\)
Divide by \(2a\)

For example, for the equation \[x^2 - 3x - 12 = 0\], with \[a = 1\], \[b = -3\], and \[c = -12\], the quadratic formula becomes:
\[ x = \frac{-(-3) \pm \root{(-3)^2 - 4(1)(-12)}}{2(1)} = \frac{3 \pm \root{9 + 48}}{2} = \frac{3 \pm 7}{2} \]
So, the roots (solutions) are: \[x = 5\text { and } x = -2\].

Example (e): For \[3x^2 - 4x - 2 = 0\], with \[a = 3\], \[b = -4\], and \[c = -2\]:
\[ x = \frac{-(-4) \pm \root{(-4)^2 - 4(3)(-2)}}{2(3)} = \frac{4 \pm \root{16 + 24}}{6} = \frac{4 \pm 8}{6} \] The solutions are: \[x = 2\text { and } x = -\frac{2}{3}\].

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