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Chapter 5: Problem 17

Solve each equation. $$ 15 x^{2}-7 x=4 $$

Short Answer

Expert verified
The solutions are \( x = \frac{4}{5} \) and \( x = -\frac{1}{3} \).

Step by step solution

01

Rewrite in Standard Form

Rewrite the equation in standard form: \( 15x^{2} - 7x - 4 = 0 \)
02

Identify Coefficients

Identify coefficients for the quadratic equation \( ax^2 + bx + c = 0 \): The coefficients are: \( a = 15 \)\( b = -7 \)\( c = -4 \)
03

Use the Quadratic Formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plug the values of \( a \), \( b \), and \( c \) into the formula:
04

Calculate the Discriminant

Calculate the discriminant using \( b^2 - 4ac \): \( (-7)^2 - 4(15)(-4) \) = \( 49 + 240 = 289 \) The discriminant \( \Delta \) is 289.
05

Solve for x

Use the quadratic formula to solve for \( x \): \( x = \frac{-(-7) \pm \sqrt{289}}{2(15)} \) \( x = \frac{7 \pm 17}{30} \) So, we get two solutions: \( x = \frac{7 + 17}{30} = \frac{24}{30} = \frac{4}{5} \) \( x = \frac{7 - 17}{30} = \frac{-10}{30} = -\frac{1}{3} \)

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Key Concepts

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

Standard Form of Quadratic Equation
When solving a quadratic equation, the first step is to write it in standard form. The standard form of a quadratic equation is denoted by:
\( ax^2 + bx + c = 0 \).
Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \).
For the equation given in the exercise:
\( 15x^2 - 7x = 4 \), we need to move all terms to one side of the equation to get:
\( 15x^2 - 7x - 4 = 0 \).
This rearranged form enables us to identify the coefficients \( a \), \( b \), and \( c \), which are essential for further steps.
Specifically, in our case:
\( a = 15 \), \( b = -7 \), and \( c = -4 \).
Quadratic Formula
With the standard form established, the next step involves using the quadratic formula. The quadratic formula is a crucial tool and is given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
This formula provides solutions to any quadratic equation.
By substituting the known values of \( a \), \( b \), and \( c \) into the formula, we can solve for \( x \).
For our equation:
\( a = 15 \), \( b = -7 \), and \( c = -4 \),
we substitute them into the formula to get:
\( x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(15)(-4)}}{2(15)} \).
Discriminant
The discriminant is a key part of the quadratic formula. It helps determine the nature of the roots of the quadratic equation.
The discriminant \( \Delta \) is given by:
\( b^2 - 4ac \).
Depending on the value of the discriminant, we can infer the following:
  • If \( \Delta > 0 \), there are two distinct real roots.
  • If \( \Delta = 0 \), there is exactly one real root.
  • If \( \Delta < 0 \), there are no real roots (the roots are complex).

For the equation we are solving, we calculate \( \Delta = (-7)^2 - 4(15)(-4) = 49 + 240 = 289 \).
Since \( \Delta = 289 \) is greater than 0, we conclude there are two distinct real roots.
These roots can be found by substituting \( \Delta \) back into the quadratic formula:
\( x = \frac{7 \pm 17}{30} \).
This gives us the solutions:
\( x = \frac{24}{30} = \frac{4}{5} \) and \( x = \frac{-10}{30} = -\frac{1}{3} \).

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