Problem 32 For Exercises 5 to $62,$ solve... [FREE SOLUTION]

Chapter 10: Problem 32

For Exercises 5 to $62,$ solve. $$6 y^{2}-19 y+15=0$$

Short Answer

Expert verified

The solutions to the equation are $y = \frac{5}{3}$ and $y = 1.5$

Step by step solution

Identify a, b, and c

For the quadratic equation $6y^{2} - 19y + 15 = 0$, identify $a$, $b$, and $c$ as follows: $a = 6$, $b = -19$, and $c = 15$

Compute the Discriminant

The discriminant is given by $D = b^{2} - 4ac$. Substituting the given values we have $D = (-19)^{2} - 4*6*15 = 361 - 360 = 1$

Apply the Quadratic Formula

Finally apply the quadratic formula, which is $y = \frac{-b \pm \sqrt{D}}{2a}$. Here $D$ is the discriminant. This gives $y = \frac{19 \pm \sqrt{1}}{12}$, resulting in the roots $y = \frac{20}{12} = \frac{5}{3}$ and $y = \frac{18}{12} = 1.5$

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Key Concepts

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

Discriminant

The discriminant is a key part of the quadratic formula. It helps in understanding how many and what type of roots a quadratic equation has. The discriminant is calculated using the formula $ D = b^2 - 4ac $, where $ a $, $ b $, and $ c $ are the coefficients of the quadratic equation $ ax^2 + bx + c = 0 $.

If $ D > 0 $, the equation has two real, distinct roots.
If $ D = 0 $, there is exactly one real root, meaning the roots are real and equal (also called repeated roots).
If $ D < 0 $, the roots are complex or imaginary because no real number solutions exist.

In our exercise, we computed the discriminant as $ D = (-19)^2 - 4 \cdot 6 \cdot 15 $. This equals $ 1 $, which is greater than zero, indicating two distinct real roots.

Quadratic Formula

The quadratic formula is a universal tool for solving any quadratic equation. This formula is particularly useful when the quadratic equation does not factor easily. The quadratic formula, $ y = \frac{-b \pm \sqrt{D}}{2a} $, gives us the solutions or roots of the quadratic equation.

Start by identifying the coefficients $ a $, $ b $, and $ c $ from the equation. In our problem, those values are $ a = 6 $, $ b = -19 $, and $ c = 15 $.
Calculate the discriminant $ D = b^2 - 4ac $.
Substitute $ b $, $ D $, and $ a $ back into the quadratic formula. This step allows you to solve for $ y $, getting the roots of the equation.

By following these steps, the quadratic formula reveals the roots, simplifying even complex quadratic equations.

Roots of Equation

The roots of a quadratic equation are the values of the variable that make the equation equal to zero. They are also sometimes referred to as solutions of the equation.
To find the roots of $ 6y^2 - 19y + 15 = 0 $, we used the quadratic formula. By substituting in $ a = 6 $, $ b = -19 $, and $ D = 1 $, we obtained two roots because the discriminant was positive:

$ y = \frac{-(-19) + \sqrt{1}}{2 \cdot 6} = \frac{20}{12} = \frac{5}{3} $
$ y = \frac{-(-19) - \sqrt{1}}{2 \cdot 6} = \frac{18}{12} = 1.5 $

These roots, $ \frac{5}{3} $ and $ 1.5 $, are the solutions to the original quadratic equation, meaning if you substitute them back into $ 6y^2 - 19y + 15 = 0 $, the equation will hold true.

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91影视

For Exercises 5 to \(62,\) solve. $$6 y^{2}-19 y+15=0$$

Short Answer

Step by step solution

Identify a, b, and c

Compute the Discriminant

Apply the Quadratic Formula

Key Concepts

Discriminant

Quadratic Formula

Roots of Equation

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