Problem 27 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by completing the square. $$10 T-5 T^{2}=4$$

Step-by-Step Solution

Answer

The solutions are \( T = 1 + \frac{\sqrt{5}}{5} \) and \( T = 1 - \frac{\sqrt{5}}{5} \).

Step 1: Rewrite in Standard Form

First, rearrange the given equation to get it into the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \). The given equation is \( 10T - 5T^2 = 4 \). Subtract 4 from both sides and rearrange to get \(-5T^2 + 10T - 4 = 0 \). Multiply the entire equation by -1 to make the leading coefficient positive: this becomes \(5T^2 - 10T + 4 = 0\).

Step 2: Move Constant to the Other Side

Keep the terms involving \(T\) on one side and move the constant term to the other side. That gives \(5T^2 - 10T = -4\).

Step 3: Normalize the Quadratic Term

To complete the square, the coefficient of \(T^2\) should be 1. So divide the entire equation by 5: \(T^2 - 2T = -\frac{4}{5}\).

Step 4: Complete the Square

To complete the square, we need to add and subtract the square of half of the coefficient of \(T\) (which is -2). Half of -2 is -1, and its square is 1. So, add and subtract 1 inside the equation: \(T^2 - 2T + 1 = -\frac{4}{5} + 1\).

Step 5: Simplify the Equation

The left side of the equation forms a perfect square: \((T - 1)^2\). Simplify the right side: \(-\frac{4}{5} + \frac{5}{5} = \frac{1}{5}\). So the equation becomes \((T - 1)^2 = \frac{1}{5}\).

Step 6: Solve for T

Take the square root of both sides.\((T - 1) = \pm \sqrt{\frac{1}{5}}\). Solving this gives \(T = 1 \pm \sqrt{\frac{1}{5}}\). Simplify further to get \(T = 1 \pm \frac{1}{\sqrt{5}}\).

Step 7: Rationalize the Denominator

To rationalize \(\frac{1}{\sqrt{5}}\), multiply both the numerator and denominator by \(\sqrt{5}\): \(\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\). The solutions become \(T = 1 + \frac{\sqrt{5}}{5}\) and \(T = 1 - \frac{\sqrt{5}}{5}\).

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial that has the highest degree of 2. This means that the variable in the equation will have a term raised to the power of 2.
Quadratic equations are generally expressed in the form:

• Standard Form: \(ax^2+bx+c=0\)
• Here, \(a, b,\) and \(c\) are constants with \(a eq 0\)
• The quadratic term \(ax^2\) represents a parabolic curve

These equations are involved in many real-life scenarios, such as calculating areas, modeling trajectories, and can be useful in physics and engineering fields. Solving quadratic equations is fundamental for progressing in algebra.

Perfect Square Trinomial

A perfect square trinomial is a special kind of quadratic expression that is particularly easy to solve once identified. It has the form:

• \((a+b)^2 = a^2 + 2ab + b^2\)
• \((a-b)^2 = a^2 - 2ab + b^2\)

To complete the square in a quadratic equation, you aim to rewrite the equation such that one side becomes a perfect square trinomial. This usually involves dividing the linear term's coefficient by two, squaring the result, and then adding and subtracting this square. For example, in the expression \(T^2 - 2T\), using half of the linear term 2 creates the perfect square trinomial \((T-1)^2\). This approach simplifies solving quadratic equations significantly.

Rationalizing the Denominator

Rationalizing the denominator is a method used in mathematics to eliminate radicals (like squareroots) from the bottom (denominator) of a fraction.
It helps in expressing the result in a more standard form, which is usually preferred:

• Multiply numerator and denominator by the conjugate of the denominator
• In this case, if the denominator was \(\sqrt{5}\), you would multiply by \(\sqrt{5}/\sqrt{5}\)
• The solution \(\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\) demonstrates a rationalized denominator

This common practice not only simplifies expressions but also makes further arithmetic operations more straightforward.

Solving Quadratic Equations

There are several methods to solve quadratic equations, each useful under different circumstances:

• Factoring, if the quadratic can be factored easily
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Completing the square, often used when other methods are cumbersome

Completing the square involves rearranging terms and creating a perfect square trinomial, allowing you to take the square root of both sides to solve for the variable. Once you've isolated the variable, you can optionally rationalize any denominator for more polished answers. Solving quadratic equations is a core skill in algebra, helping students apply their knowledge to solve a variety of mathematical problems.