Problem 43 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ 6 w+7 w^{1 / 2}-20=0 $$

Step-by-Step Solution

Answer

The solution is \( w = \frac{16}{9} \) and \( w = \frac{25}{4} \) (requires verification for original equation matching).

Step 1: Identify the Equation Type

The given equation is a polynomial equation with terms involving different powers of \( w \): \( 6w \), \( 7w^{1/2} \), and a constant term \(-20\). Note that \( w^{1/2} \) is equivalent to the square root of \( w \). This suggests using substitution to simplify the equation.

Step 2: Introduce a Substitution

Let \( x = w^{1/2} \), which means \( x^2 = w \). Rewrite the equation using this substitution: \( 6x^2 + 7x - 20 = 0 \). This transforms the original equation into a quadratic equation in terms of \( x \).

Step 3: Solve the Quadratic Equation

The quadratic equation \( 6x^2 + 7x - 20 = 0 \) can be solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6 \), \( b = 7 \), and \( c = -20 \).

Step 4: Calculate the Discriminant

Calculate the discriminant \( \Delta = b^2 - 4ac = 7^2 - 4(6)(-20) = 49 + 480 = 529 \). Since \( \Delta > 0 \), there are two real solutions for \( x \).

Step 5: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \[ x = \frac{-7 \pm \sqrt{529}}{12} \] \[ x = \frac{-7 \pm 23}{12} \]This results in two possible solutions for \( x \): 1. \( x = \frac{16}{12} = \frac{4}{3} \) 2. \( x = \frac{-30}{12} = -\frac{5}{2} \)

Step 6: Express Solutions in Terms of w

Recall that \( x = w^{1/2} \). For each solution of \( x \):1. \( x = \frac{4}{3} \) implies \( w = (\frac{4}{3})^2 = \frac{16}{9} \).2. \( x = -\frac{5}{2} \) implies \( w = (-\frac{5}{2})^2 = \frac{25}{4} \) (this value is mathematically valid for the square root but negative values aren’t meaningful for the original substitution context since the square root of a negative number isn’t real).

Step 7: Verify Solutions

Substitute \( w = \frac{16}{9} \) back into the original equation to verify it satisfies: \( 6(\frac{16}{9}) + 7(\frac{16}{9})^{1/2} - 20 = 0 \). Similarly, check for \( w = \frac{25}{4} \). Both solutions need to solve the original equation for true verification.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used for solving quadratic equations. These are equations of the form \( ax^2 + bx + c = 0 \). The quadratic formula looks like this:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps find the roots, or solutions, of the quadratic equation by determining the values of \( x \) that make the equation true. To apply it, you need to identify the coefficients \( a \), \( b \), and \( c \) from the equation. Here:

• \( a \) is the coefficient of \( x^2 \),
• \( b \) is the coefficient of \( x \),
• and \( c \) is the constant term.

In our example, after substituting the original polynomial with something more manageable, we solve \( 6x^2 + 7x - 20 = 0 \) using this formula, where \( a = 6 \), \( b = 7 \), and \( c = -20 \). By substituting these into the formula, we find the values of \( x \) that satisfy the equation, giving us the possible solutions for \( x \). Remember, the quadratic formula is reliable as long as you correctly plug in the values and simplify step by step!
This is particularly useful because it works all the time for any quadratic equation, even when factoring is impossible.

Substitution Method

The substitution method is an effective strategy for solving equations that aren't quite traditional quadratics initially. In this problem, we dealt with a term \( w^{1/2} \), also known as the square root of \( w \). This creates challenges when attempting to factor or apply regular methods directy.
The idea behind substitution involves transforming the original equation into a simpler, more recognizable form, often a quadratic one. This is done by substituting a variable for an expression in the equation. In this exercise:

• Let \( x = w^{1/2} \), which implies \( x^2 = w \).
• Substitute these into the original equation to get \( 6x^2 + 7x - 20 = 0 \).

This technique effectively converts a complex-looking equation into a standard quadratic form, which we can then solve using established methods like the quadratic formula. Once solved for \( x \), we revert back to terms of \( w \) to find the solutions for the original problem. Substitution not only simplifies the process but also helps avoid errors that might arise when dealing with exponents or roots directly.

Discriminant

The discriminant is a significant component of the quadratic formula that gives insight into the nature of the roots of a quadratic equation. It is the part under the square root sign in the quadratic formula, calculated as \( \Delta = b^2 - 4ac \).
The value of the discriminant determines the type and number of solutions you can expect:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (also known as a repeated or double root).
• If \( \Delta < 0 \), there are no real roots; instead, the equation has two complex roots.

In the example provided, the calculated discriminant is \( 529 \), which is greater than zero. This indicates the presence of two distinct real solutions for \( x \). This aspect of the quadratic formula not only tells us how many roots to expect but also predicts whether the solutions will be real numbers or involve complex numbers. Understanding and calculating the discriminant helps in foreseeing the nature of solutions, saving unnecessary computations and providing clearer outlook on the solutions properties.