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Magazine Chapter 7: Problem 27
Solve the given quadratic equations by completing the square. $$10 T-5 T^{2}=4$$
Short Answer
Expert verified
The solutions are \( T = 1 + \frac{\sqrt{5}}{5} \) and \( T = 1 - \frac{\sqrt{5}}{5} \).
Step by step solution
01
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 \(5T^2 - 10T + 4 = 0\).
02
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 \(5T^2 - 10T = -4\).
03
Normalize the Quadratic Term
To complete the square, the coefficient of \(T^2\) should be 1. So divide the entire equation by 5: \(T^2 - 2T = -\frac{4}{5}\).
04
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: \(T^2 - 2T + 1 = -\frac{4}{5} + 1\).
05
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 \((T - 1)^2 = \frac{1}{5}\).
06
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 \(T = 1 \pm \frac{1}{\sqrt{5}}\).
07
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 \(T = 1 - \frac{\sqrt{5}}{5}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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
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:
- \((a+b)^2 = a^2 + 2ab + b^2\)
- \((a-b)^2 = a^2 - 2ab + b^2\)
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:
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
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
