Simplifying Expressions And Solving Quadratic Equations
This section aims to simplify a complex fraction involving mixed numbers and division. Simplifying complex fractions requires a strong understanding of fraction arithmetic, including converting mixed numbers to improper fractions, finding common denominators, and applying the order of operations (PEMDAS/BODMAS). The key to success lies in breaking down the problem into manageable steps, ensuring accuracy at each stage. Let's dive into the step-by-step solution, highlighting the fundamental principles that govern fraction manipulation and aiming to provide a clear and concise approach to tackle similar mathematical challenges. First, convert the mixed numbers into improper fractions. This involves multiplying the whole number part by the denominator and adding the numerator, then placing the result over the original denominator. So, $6 \frac2}{3}$ becomes $\frac{(6 \times 3) + 2}{3} = \frac{20}{3}$, $3 \frac{4}{15}$ becomes $\frac{(3 \times 15) + 4}{15} = \frac{49}{15}$, and $1 \frac{3}{5}$ becomes $\frac{(1 \times 5) + 3}{5} = \frac{8}{5}$. The expression now looks like $\frac{20}{3} \div\left(\frac{49}{15} - \frac{8}{5}\right)$. Next, we need to handle the subtraction within the parentheses. To subtract fractions, they must have a common denominator. The least common multiple (LCM) of 15 and 5 is 15. Convert $\frac{8}{5}$ to an equivalent fraction with a denominator of 15 by multiplying both the numerator and the denominator by 35} \times \frac{3}{3} = \frac{24}{15}$. Now the expression inside the parentheses is $\frac{49}{15} - \frac{24}{15}$, which simplifies to $\frac{49 - 24}{15} = \frac{25}{15}$. This fraction can be further simplified by dividing both the numerator and the denominator by their greatest common divisor (GCD), which is 515} = \frac{25 \div 5}{15 \div 5} = \frac{5}{3}$. Now the original expression is reduced to $\frac{20}{3} \div \frac{5}{3}$. Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{5}{3}$ is $\frac{3}{5}$. Therefore, the expression becomes $\frac{20}{3} \times \frac{3}{5}$. Multiply the numerators together and the denominators together3 \times 5} = \frac{60}{15}$. Finally, simplify the fraction by dividing both the numerator and the denominator by their GCD, which is 15{15} = \frac{60 \div 15}{15 \div 15} = 4$. Thus, the simplified form of the given expression is 4. This process demonstrates the importance of a systematic approach when dealing with complex fractions, ensuring accuracy and clarity in each step. By breaking down the problem into smaller, more manageable parts, the overall simplification becomes significantly easier.
Answer: 4
This part involves solving a quadratic equation, a fundamental topic in algebra. Quadratic equations are polynomial equations of the second degree, generally represented in the form $ax^2 + bx + c = 0$, where a, b, and c are constants, and a is not equal to zero. The solutions to a quadratic equation, also known as roots, are the values of x that satisfy the equation. These roots can be real or complex numbers and can be found using various methods such as factoring, completing the square, or the quadratic formula. In this specific problem, we are given that $x = -\frac2}{3}$ is a solution to the quadratic equation $px^2 + 16x + 4 = 0$. This information is crucial because it allows us to determine the value of the unknown coefficient 'p'. By substituting the given value of x into the equation, we can create a linear equation in terms of 'p', which can then be easily solved. Once we find the value of 'p', the equation is fully defined, and we can proceed to find the other solution (root) of the quadratic equation. This usually involves employing techniques such as factoring, using the quadratic formula, or applying Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. The approach we take to solve this problem illustrates a common strategy in algebra3}$ is a solution to the equation $px^2 + 16x + 4 = 0$, we can substitute this value into the equation3}\right)^2 + 16\left(-\frac{2}{3}\right) + 4 = 0$. Simplifying the equation, we get $p\left(\frac{4}{9}\right) - \frac{32}{3} + 4 = 0$. To eliminate the fractions, we can multiply the entire equation by 9{9}\right) - \frac{32}{3} + 4\right] = 9 \times 0$, which gives us $4p - 96 + 36 = 0$. Combining the constants, we have $4p - 60 = 0$. Adding 60 to both sides, we get $4p = 60$. Dividing both sides by 4, we find $p = 15$. Therefore, the value of p is 15. This step demonstrates the power of substitution in solving algebraic problems. By replacing the variable x with its known value, we transformed the quadratic equation into a linear equation that was easily solvable for the unknown coefficient p.
i) the value of $p$
Answer: p = 15
ii) the value of $x$ which
In part ii), we need to find the other value of x that satisfies the quadratic equation. Now that we know $p = 15$, our equation becomes $15x^2 + 16x + 4 = 0$. To find the roots of this quadratic equation, we can use several methods, including factoring, completing the square, or applying the quadratic formula. Factoring is often the quickest method if the quadratic expression can be factored easily. Let's try factoring the quadratic expression. We are looking for two binomials of the form $(ax + b)(cx + d)$ such that when multiplied, they give us $15x^2 + 16x + 4$. The product of the first terms should be $15x^2$, and the product of the last terms should be 4. The sum of the outer and inner products should be 16x. After some trial and error, we find that the expression can be factored as $(3x + 2)(5x + 2) = 0$. To verify this, we can expand the factored form: $(3x + 2)(5x + 2) = 15x^2 + 6x + 10x + 4 = 15x^2 + 16x + 4$, which matches our original quadratic equation. Now that we have factored the equation, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for x: $3x + 2 = 0$ or $5x + 2 = 0$. Solving $3x + 2 = 0$, we subtract 2 from both sides to get $3x = -2$, and then divide by 3 to find $x = -\frac{2}{3}$. This is the solution we were given initially. Solving $5x + 2 = 0$, we subtract 2 from both sides to get $5x = -2$, and then divide by 5 to find $x = -\frac{2}{5}$. Therefore, the other value of x that satisfies the equation is $x = -\frac{2}{5}$. This solution demonstrates the utility of factoring in solving quadratic equations. By expressing the quadratic expression as a product of two linear factors, we were able to easily identify the roots of the equation. This is a powerful technique that is widely used in algebra and calculus.
Answer: $x = -\frac{2}{5}$
