If $\alpha$ And $\beta$ Are Zeroes Of The Quadratic Polynomial $2x^2 + 7x + 5$, Find The Value Of $\alpha^2 + \beta^2 + \alpha \beta$.

If $\alpha$ and $\beta$ are zeroes of the quadratic polynomial $2x^2 + 7x + 5$, find the value of $\alpha^2 + \beta^2 + \alpha \beta$

In this article, we will explore the relationship between the zeroes of a quadratic polynomial and the value of a specific expression involving these zeroes. We will use the given quadratic polynomial $2x^2 + 7x + 5$ and its zeroes $\alpha$ and $\beta$ to find the value of $\alpha^2 + \beta^2 + \alpha \beta$.

The Relationship Between Zeroes and Coefficients

The quadratic polynomial $2x^2 + 7x + 5$ can be written in the form $ax^2 + bx + c$, where $a = 2$, $b = 7$, and $c = 5$. The zeroes of this polynomial are $\alpha$ and $\beta$, which satisfy the equation $2x^2 + 7x + 5 = 0$. We can use Vieta's formulas to relate the zeroes to the coefficients of the polynomial.

Vieta's Formulas

Vieta's formulas state that for a quadratic polynomial $ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$, the following relationships hold:

• $\alpha + \beta = -\frac{b}{a}$
• $\alpha \beta = \frac{c}{a}$

In this case, we have:

• $\alpha + \beta = -\frac{7}{2}$
• $\alpha \beta = \frac{5}{2}$

Finding the Value of $\alpha^2 + \beta^2 + \alpha \beta$

We are asked to find the value of $\alpha^2 + \beta^2 + \alpha \beta$. We can start by squaring the equation $\alpha + \beta = -\frac{7}{2}$:

$$\alpha^2 + 2\alpha \beta + \beta^2 = \frac{49}{4}$$

Now, we can substitute the value of $\alpha \beta = \frac{5}{2}$ into this equation:

$$\alpha^2 + 2\left(\frac{5}{2}\right) + \beta^2 = \frac{49}{4}$$

Simplifying this equation, we get:

$$\alpha^2 + \beta^2 + 5 = \frac{49}{4}$$

Subtracting 5 from both sides, we get:

$$\alpha^2 + \beta^2 = \frac{49}{4} - 5$$

Simplifying this expression, we get:

$$\alpha^2 + \beta^2 = \frac{49 - 20}{4}$$

$$\alpha^2 + \beta^2 = \frac{29}{4}$$

Finally, we can substitute the value of $\alpha \beta = \frac{5}{2}$ into the original expression:

$$\alpha^2 + \beta^2 + \alpha \beta = \frac{29}{4} + \frac{5}{2}$$

Simplifying this expression, we get:

$$\alpha^2 + \beta^2 + \alpha \beta = \frac{29}{4} + \frac{10}{4}$$

$$\alpha^2 + \beta^2 + \alpha \beta = \frac{39}{4}$$

In this article, we used Vieta's formulas to relate the zeroes of a quadratic polynomial to its coefficients. We then used these relationships to find the value of $\alpha^2 + \beta^2 + \alpha \beta$. We started by squaring the equation $\alpha + \beta = -\frac{7}{2}$ and substituting the value of $\alpha \beta = \frac{5}{2}$. We then simplified the resulting equation to find the value of $\alpha^2 + \beta^2$. Finally, we substituted the value of $\alpha \beta = \frac{5}{2}$ into the original expression to find the final answer.

The final answer is $\boxed{\frac{39}{4}}$.
Q&A: If $\alpha$ and $\beta$ are zeroes of the quadratic polynomial $2x^2 + 7x + 5$, find the value of $\alpha^2 + \beta^2 + \alpha \beta$

In our previous article, we explored the relationship between the zeroes of a quadratic polynomial and the value of a specific expression involving these zeroes. We used the given quadratic polynomial $2x^2 + 7x + 5$ and its zeroes $\alpha$ and $\beta$ to find the value of $\alpha^2 + \beta^2 + \alpha \beta$. In this article, we will answer some common questions related to this topic.

Q: What are the zeroes of a quadratic polynomial?

A: The zeroes of a quadratic polynomial are the values of $x$ that satisfy the equation $ax^2 + bx + c = 0$. In other words, they are the solutions to the equation.

Q: How do you find the zeroes of a quadratic polynomial?

A: To find the zeroes of a quadratic polynomial, you can use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

This formula will give you two solutions, which are the zeroes of the polynomial.

Q: What is Vieta's formulas?

A: Vieta's formulas are a set of relationships between the coefficients of a quadratic polynomial and its zeroes. They state that for a quadratic polynomial $ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$, the following relationships hold:

• $\alpha + \beta = -\frac{b}{a}$
• $\alpha \beta = \frac{c}{a}$

Q: How do you use Vieta's formulas to find the value of $\alpha^2 + \beta^2 + \alpha \beta$?

A: To find the value of $\alpha^2 + \beta^2 + \alpha \beta$, you can start by squaring the equation $\alpha + \beta = -\frac{b}{a}$. This will give you an equation involving $\alpha^2$, $\beta^2$, and $\alpha \beta$. You can then substitute the value of $\alpha \beta = \frac{c}{a}$ into this equation and simplify to find the final answer.

Q: What is the final answer to the problem?

A: The final answer to the problem is $\boxed{\frac{39}{4}}$.

Q: Can you provide an example of how to use Vieta's formulas to find the value of $\alpha^2 + \beta^2 + \alpha \beta$?

A: Let's consider the quadratic polynomial $x^2 + 5x + 6$. We can use Vieta's formulas to find the value of $\alpha^2 + \beta^2 + \alpha \beta$.

First, we can find the zeroes of the polynomial using the quadratic formula:

$$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$$

Simplifying this expression, we get:

$$x = \frac{-5 \pm \sqrt{1}}{2}$$

This gives us two solutions, $\alpha = -3$ and $\beta = -2$.

Next, we can use Vieta's formulas to find the value of $\alpha^2 + \beta^2 + \alpha \beta$:

• $\alpha + \beta = -3 - 2 = -5$
• $\alpha \beta = (-3)(-2) = 6$

Now, we can square the equation $\alpha + \beta = -5$:

$$\alpha^2 + 2\alpha \beta + \beta^2 = 25$$

Substituting the value of $\alpha \beta = 6$ into this equation, we get:

$$\alpha^2 + 2(6) + \beta^2 = 25$$

Simplifying this expression, we get:

$$\alpha^2 + \beta^2 = 25 - 12$$

$$\alpha^2 + \beta^2 = 13$$

Finally, we can substitute the value of $\alpha \beta = 6$ into the original expression:

$$\alpha^2 + \beta^2 + \alpha \beta = 13 + 6$$

Simplifying this expression, we get:

$$\alpha^2 + \beta^2 + \alpha \beta = 19$$

Therefore, the final answer to the problem is $\boxed{19}$.

In this article, we answered some common questions related to the problem of finding the value of $\alpha^2 + \beta^2 + \alpha \beta$ using Vieta's formulas. We provided examples and explanations to help illustrate the concepts and procedures involved. We hope that this article has been helpful in clarifying any doubts or questions you may have had about this topic.