Problem 64 - Elementary Algebra

Question

(a) solve. (b) check. $$ 6 v^{2}+31 v+35=0 $$

Step-by-Step Solution

Answer

The solutions are \(v = -\frac{5}{3}\) and \(v = -3.5\).

Step 1: Identify coefficients

Here, we have a quadratic equation of the form ax^2 + bx + c = 0a = 6, b = 31, c = 35

Step 2: Calculate the discriminant

The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by \(D = b^2 - 4ac\)Substituting the values, we get \(D = 31^2 - 4(6)(35)\)\(D = 961 - 840 = 121\)

Step 3: Find the roots using the quadratic formula

The roots of the quadratic equation ax^2 + bx + c = 0 can be found using \(x = \frac{-b \pm \sqrt{D}}{2a}\)Since D > 0, there will be two distinct real roots. Substituting the values, we get \(v = \frac{-31 \pm 11}{12}\)So,\(v_1 = \frac{-31 + 11}{12} = \frac{-20}{12} = \frac{-5}{3}\)\(v_2 = \frac{-31 - 11}{12} = \frac{-42}{12} = -3.5\)

Step 4: Check the roots

Substitute \(v_1 = -\frac{5}{3}\) and \(v_2 = -3.5\) back into the original equation to verify the solutions: \(6\left(-\frac{5}{3}\right)^2 + 31\left(-\frac{5}{3}\right) + 35 = 0\)\(6\left(\frac{25}{9}\right) - \left(\frac{155}{3}\right) + 35 = 0\)\(\frac{150}{9} - \frac{465}{9} + \frac{315}{9} = 0\)\(\frac{150 - 465 + 315}{9} = 0\)\(0 = 0\)and\(6(-3.5)^2 + 31(-3.5) + 35 = 0\)\(6(12.25) - 108.5 + 35 = 0\)\(73.5 - 108.5 + 35 = 0\)\(0 = 0\)

Key Concepts

quadratic formula

The quadratic formula is a powerful tool used to solve quadratic equations. A quadratic equation is any equation that can be written in the form: \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are coefficients, which are just constants. The quadratic formula gives us a way to find the values of \(x\) that make the equation true. It is written as:

\[x = \frac{-b \pm \sqrt{D}}{2a}\]

In this formula, \(D\) stands for the discriminant, which we will talk about in the next section. The symbol \(\pm\) means that there will usually be two solutions: one that uses a plus sign and one that uses a minus sign. The quadratic formula can solve any quadratic equation, even if it doesn’t factor nicely. By substituting the correct values for \(a\), \(b\), and \(c\), we can find the solutions \(v_1\) and \(v_2\).

• Start with your quadratic equation, for example, \(6v^2 + 31v + 35 = 0\).
• Identify coefficients: Here, \(a = 6\), \(b = 31\), and \(c = 35\).
• Substitute these values into the quadratic formula.
• Solve for the two values of \(v\).

discriminant

The discriminant is an important part of the quadratic formula and it helps us determine the nature of the solutions to a quadratic equation. It is found inside the square root symbol in the quadratic formula. The discriminant is given by:

\[D = b^2 - 4ac\]

Depending on the value of \(D\), we can tell if the quadratic equation has:

• Two distinct real solutions (when \(D > 0\))
• One real solution (when \(D = 0\))
• No real solutions (when \(D < 0\))

For the equation \(6v^2 + 31v + 35 = 0\), we substitute \(a = 6\), \(b = 31\), and \(c = 35\) into the discriminant formula:

\[D = 31^2 - 4(6)(35) = 961 - 840 = 121\]

Since \(D = 121\) which is greater than 0, it means we have two distinct real solutions.

verifying solutions

After finding the solutions of a quadratic equation using the quadratic formula, it's important to verify them. This ensures that our solutions are correct. Verifying solutions involves substituting them back into the original equation to see if they satisfy it. Here are the steps:

• Substitute each root back into the original equation.
• Simplify the expression.
• Check if the left-hand side equals the right-hand side.

For our solutions:
Root \(v_1 = -\frac{5}{3}\):
\[6\left( -\frac{5}{3} \right)^2 + 31\left( -\frac{5}{3} \right) + 35 = 0\]
Simplifying, we get:
\[6\left( \frac{25}{9} \right) - \left( \frac{155}{3} \right) + 35 = 0\]
which simplifies to:
\[\frac{150}{9} - \frac{465}{9} + \frac{315}{9} = 0\]
Finally: \[\frac{0}{9} = 0\]
Root \(v_2 = -3.5\):
\[6(-3.5)^2 + 31(-3.5) + 35 = 0\]
Simplifying, we get:
\[6(12.25) - 108.5 + 35 = 0\]
which simplifies to:
\[73.5 - 108.5 + 35 = 0\]
Finally: \[0 = 0\]
Both roots satisfy the original equation, so our solutions are correct!