Finding Quadratic Equations With Specific Solutions A Step-by-Step Guide

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Understanding quadratic equations and their solutions is a fundamental concept in algebra. This article delves into the process of identifying the quadratic equation that corresponds to the given solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$. We will explore the quadratic formula, its derivation, and how to use it effectively to solve and verify quadratic equations. Furthermore, we will analyze the relationship between the solutions and the coefficients of the quadratic equation. This exploration will not only help in solving this particular problem but also strengthen the understanding of quadratic equations in general.

Decoding the Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions (or roots) of any quadratic equation in the standard form $ax^2 + bx + c = 0$, where a, b, and c are constants and $a \ne 0$. The formula is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula provides two possible solutions, one with a plus sign and the other with a minus sign, representing the two roots of the quadratic equation. The expression inside the square root, $b^2 - 4ac$, is known as the discriminant, which determines the nature of the roots. If the discriminant is positive, there are two distinct real roots; if it is zero, there is exactly one real root (a repeated root); and if it is negative, there are two complex roots.

The derivation of the quadratic formula involves completing the square on the general quadratic equation $ax^2 + bx + c = 0$. First, divide the entire equation by a:

x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0

Next, move the constant term to the right side of the equation:

x2+bax=βˆ’cax^2 + \frac{b}{a}x = -\frac{c}{a}

To complete the square, add $\left(\frac{b}{2a}\right)^2 = \frac{b2}{4a2}$ to both sides:

x2+bax+b24a2=βˆ’ca+b24a2x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}

Rewrite the left side as a perfect square and simplify the right side:

(x+b2a)2=b2βˆ’4ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Take the square root of both sides:

x+b2a=Β±b2βˆ’4ac2ax + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}

Finally, isolate x to obtain the quadratic formula:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This derivation highlights the mathematical foundation of the formula and provides a deeper understanding of its application. The quadratic formula is not just a memorized equation; it's a result of logical steps in completing the square, making it a versatile tool for solving quadratic equations.

Applying the Quadratic Formula to Find the Equation

Our primary goal is to identify which of the given quadratic equations has the solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$. To achieve this, we will reverse-engineer the quadratic formula. We are given the solutions, and we need to find the coefficients a, b, and c of the quadratic equation $ax^2 + bx + c = 0$. By comparing the given solutions with the quadratic formula, we can deduce the values of these coefficients. This process involves careful observation and algebraic manipulation.

Comparing the given solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$ with the general form of the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we can make the following observations:

  1. The term outside the square root in the numerator is 5, which corresponds to $-b$. However, we also have the denominator 3, which corresponds to $2a$. Thus, we can infer that $-b = 5$ and $2a = 3$, which gives us $b = -5$ and $a = \frac{3}{2}$. But since we prefer integer coefficients in a quadratic equation, we will double the entire equation later to clear the fraction. For now, we keep $a = \frac{3}{2}$. Alternatively, we can directly compare $2a = 3$, which gives us $a = \frac{3}{2}$.
  2. The term inside the square root is $(2\sqrt{7})^2 = 4 \times 7 = 28$. This corresponds to the discriminant $b^2 - 4ac$. Substituting $b = -5$, we get $(-5)^2 - 4ac = 25 - 4ac$. Thus, we have $25 - 4ac = 28$. Substituting $a = \frac{3}{2}$, the equation becomes $25 - 4(\frac{3}{2})c = 28$, which simplifies to $25 - 6c = 28$. Solving for c, we get $-6c = 3$, so $c = -\frac{1}{2}$.
  3. To avoid fractional coefficients, we multiply the entire equation by 2. Thus, we have $a = 3$, $b = -10$, and $b^2 - 4ac$ term must yield $28 * 2 * 2 = 112$.

Using $b = -10$, then we have $(-10)^2 - 4(3)c = 100 - 12c$ which must be equal to $28 * 4 = 112$. Thus $100 - 12c = 112$, solving for c gives $-12c = 12$, so $c = -1$.

Therefore, our quadratic equation is $3x^2 - 10x - 1 = 0$. To confirm, we can substitute these values into the quadratic formula:

x=βˆ’(βˆ’10)Β±(βˆ’10)2βˆ’4(3)(βˆ’1)2(3)x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(-1)}}{2(3)}

x=10Β±100+126x = \frac{10 \pm \sqrt{100 + 12}}{6}

x=10Β±1126x = \frac{10 \pm \sqrt{112}}{6}

x=10Β±16Γ—76x = \frac{10 \pm \sqrt{16 \times 7}}{6}

x=10Β±476x = \frac{10 \pm 4\sqrt{7}}{6}

x=5Β±273x = \frac{5 \pm 2\sqrt{7}}{3}

This matches the given solutions, confirming that the quadratic equation $3x^2 - 10x - 1 = 0$ has the solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$. This methodical approach demonstrates how the quadratic formula can be used not only to find solutions but also to identify the equation given its solutions.

Analyzing the Given Options

Now that we have determined the correct quadratic equation, let's analyze the given options to understand why the others are incorrect. This will further solidify our understanding of the relationship between a quadratic equation and its solutions. We can apply the quadratic formula to each option and compare the results with the given solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$.

  1. 3x^2 - 10x - 1 = 0$: As we have already shown, this equation yields the correct solutions.

  2. 3x^2 - 5x + 7 = 0$: Applying the **quadratic formula**, we get: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(7)}}{2(3)}

    x=5Β±25βˆ’846x = \frac{5 \pm \sqrt{25 - 84}}{6}

    x=5Β±βˆ’596x = \frac{5 \pm \sqrt{-59}}{6}

    These solutions are complex numbers, which do not match the given solutions.

  3. 3x^2 - 5x - 1 = 0$: Applying the **quadratic formula**, we get: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-1)}}{2(3)}

    x=5Β±25+126x = \frac{5 \pm \sqrt{25 + 12}}{6}

    x=5Β±376x = \frac{5 \pm \sqrt{37}}{6}

    These solutions do not match the given solutions.

  4. 3x^2 - 10x + 6 = 0$: Applying the **quadratic formula**, we get: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(6)}}{2(3)}

    x=10Β±100βˆ’726x = \frac{10 \pm \sqrt{100 - 72}}{6}

    x=10Β±286x = \frac{10 \pm \sqrt{28}}{6}

    x=10Β±276x = \frac{10 \pm 2\sqrt{7}}{6}

    x=5Β±73x = \frac{5 \pm \sqrt{7}}{3}

    These solutions are similar but not identical to the given solutions. The difference lies in the term inside the square root.

This analysis demonstrates the importance of carefully applying the quadratic formula and comparing the results. Each quadratic equation has a unique set of solutions, and even a slight change in the coefficients can lead to different roots. The discriminant, $b^2 - 4ac$, plays a crucial role in determining the nature and form of the solutions.

The Significance of the Discriminant

The discriminant, $b^2 - 4ac$, is a key component of the quadratic formula and provides valuable information about the nature of the roots of a quadratic equation. As mentioned earlier, the discriminant determines whether the roots are real and distinct, real and equal, or complex. Understanding the discriminant can save time and effort in solving quadratic equations, as it gives a quick preview of the type of solutions to expect.

  1. If $b^2 - 4ac > 0$, the quadratic equation has two distinct real roots. This means the parabola represented by the quadratic equation intersects the x-axis at two different points.
  2. If $b^2 - 4ac = 0$, the quadratic equation has exactly one real root (a repeated root). This means the parabola touches the x-axis at exactly one point, the vertex.
  3. If $b^2 - 4ac < 0$, the quadratic equation has two complex roots. This means the parabola does not intersect the x-axis.

In our problem, the solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$ are real and distinct. This implies that the discriminant of the correct quadratic equation should be positive. Let's verify this for the correct equation, $3x^2 - 10x - 1 = 0$. The discriminant is:

(βˆ’10)2βˆ’4(3)(βˆ’1)=100+12=112(-10)^2 - 4(3)(-1) = 100 + 12 = 112

Since $112 > 0$, this confirms that the equation has two distinct real roots. For the other options, we saw that the discriminant was negative for $3x^2 - 5x + 7 = 0$, resulting in complex roots, and different for the other options, leading to different real roots.

Understanding the discriminant not only helps in solving quadratic equations but also in visualizing the graphs of quadratic functions. The sign of the discriminant is directly related to the number of x-intercepts of the parabola. This connection between algebra and geometry provides a more complete understanding of quadratic equations and their applications.

Conclusion

In summary, we have successfully identified the quadratic equation $3x^2 - 10x - 1 = 0$ as the one with solutions $x = \frac{5 \pm 2\sqrt{7}}{3}$. This was achieved by reverse-engineering the quadratic formula, comparing the given solutions with the general form of the formula, and deducing the coefficients of the quadratic equation. We also analyzed the incorrect options to understand why they did not match the given solutions and reinforced the importance of the discriminant in determining the nature of the roots.

This exploration underscores the significance of the quadratic formula as a fundamental tool in algebra. It allows us to solve any quadratic equation, regardless of the complexity of its coefficients. Furthermore, the process of finding a quadratic equation given its solutions enhances our understanding of the relationship between the roots and the coefficients. The discriminant provides valuable insight into the nature of the roots, connecting algebraic solutions with geometric interpretations.

By mastering the quadratic formula and related concepts, we can confidently tackle a wide range of quadratic equation problems and gain a deeper appreciation for the elegance and power of algebra. The ability to solve and analyze quadratic equations is essential not only in mathematics but also in various fields such as physics, engineering, and computer science, where quadratic relationships often arise.