Problem 35 - Precalculus

Question

Use the quadratic formula to solve each equation. In Exercises \(34-39,\) give two forms for each solution: an expression containing a radical and a calculator approximation rounded off to two decimal places. $$4 x^{2}-3 x-9=0$$

Step-by-Step Solution

Answer

Exact: \( x = \frac{3 \pm \sqrt{153}}{8} \); Approx: \( x \approx 1.92, -1.17 \).

Step 1: Identify the coefficients

For the quadratic equation in the form of \(ax^2 + bx + c = 0\), identify the coefficients \(a = 4\), \(b = -3\), and \(c = -9\).

Step 2: Write the quadratic formula

The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 3: Calculate the discriminant

Substitute the values of \(a\), \(b\), and \(c\) into the discriminant formula: \(b^2 - 4ac\).\( (-3)^2 - 4 \cdot 4 \cdot (-9) = 9 + 144 = 153 \).

Step 4: Compute the square root

Find the square root of the discriminant:\[ \sqrt{153} \approx 12.369 \]

Step 5: Substitute into the quadratic formula

Substitute \(b = -3\), \(a = 4\), and \(\sqrt{153}\) into the quadratic formula:\[ x = \frac{3 \pm 12.369}{8} \]

Step 6: Find the exact solutions

Using the radical:\[ x = \frac{3 \pm \sqrt{153}}{8} \]This gives the two solutions \( x = \frac{3 + \sqrt{153}}{8} \) and \( x = \frac{3 - \sqrt{153}}{8} \).

Step 7: Calculate the approximate solutions

Using a calculator, compute the decimal approximations:- For \( x = \frac{3 + \sqrt{153}}{8} \), the approximation is \( x \approx 1.92 \).- For \( x = \frac{3 - \sqrt{153}}{8} \), the approximation is \( x \approx -1.17 \).

Key Concepts

Discriminant

In quadratic equations, the discriminant is a super important component. It helps us figure out how many and what type of solutions a quadratic equation has. The discriminant is found inside the quadratic formula, and it is given by the expression: \( b^2 - 4ac \). This piece of magic can let us know whether our quadratic equation will have:

• Two real and distinct solutions when the discriminant is positive.
• One real, repeated solution when the discriminant is zero.
• No real solutions (but two complex ones) when the discriminant is negative.

In our problem, we calculated the discriminant as 153. Since 153 is positive, it tells us we have two different real solutions. A positive discriminant is always a good sign for a variety of beautiful solutions!

Quadratic Equation

A quadratic equation is any equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). It’s the key feature that gives us the classic parabola shape in graphs. Nearly anything that curves like a U or an upside-down U can often be modelled by a quadratic equation.

Let's look at our equation: \( 4x^2 - 3x - 9 = 0 \). Notice how we can identify:

• \( a = 4 \)
• \( b = -3 \)
• \( c = -9 \)

From there, we apply these coefficients in the quadratic formula to find the solutions. Adjusting the parameters \( a, b, \) and \( c \) shifts and stretches (or squishes) the parabola. Relative to its graph, changing the sign of \( a \) flips it, \( b \) shifts it horizontally, and \( c \) moves it up or down.

Radical Expressions

Radical expressions play a major role in algebra and are key parts of the quadratic formula. The word "radical" really just refers to square roots and higher-level roots expressed with the radical symbol \( \sqrt{} \). When we talk about radical expressions, you should be thinking of anything that involves this symbol.

In the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), the \( \sqrt{b^2 - 4ac} \) part is our radical expression. In our specific case, it becomes \( \sqrt{153} \). This exact radical form gives us a precise, as-inexact-as-possible solution before rounding. Feeling comfortable with radicals also helps demystify a lot of algebraic tasks.

Exact Solution

Exact solutions in mathematics aim to provide the most accurate answer possible without approximations. When using the quadratic formula, exact solutions often involve radicals, showing us precisely where the solutions are located with respect to each other.

In our equation \( 4x^2 - 3x - 9 = 0 \), we found the exact solutions as \( x = \frac{3 + \sqrt{153}}{8} \) and \( x = \frac{3 - \sqrt{153}}{8} \). Having these exact solutions gives us a full picture of the equation's landscape, revealing the x-intercepts on the graph with pinpoint accuracy.

• \( x = \frac{3 + \sqrt{153}}{8} \)
• \( x = \frac{3 - \sqrt{153}}{8} \)

Even if it is a bit intimidating, exact solutions help maintain mathematical integrity across calculations.

Calculator Approximation

Calculator approximations allow us to convert exact solutions, involving tricky radicals or fractions, into easy-to-digest decimal numbers. This is particularly useful in contexts where a rough, more intuitive understanding of a solution is required.

For our equation \( 4x^2 - 3x - 9 = 0 \), we used a calculator to approximate our solutions to two decimal places:

• \( x \approx 1.92 \) for \( x = \frac{3 + \sqrt{153}}{8} \)
• \( x \approx -1.17 \) for \( x = \frac{3 - \sqrt{153}}{8} \)

By rounding to two decimal places, we make the solutions much more approachable while sacrificing just a little bit of precision. Calculator approximations are everyday tools in science and engineering, where a full-blown, super-exact answer isn't needed.