Solving Quadratic Equations By Completing The Square
TLDRThis lesson teaches how to solve quadratic equations by completing the square. It walks through three examples, starting with a simple equation x^2 + 4x = 5, leading to solutions x = 1 and x = -5. The second example involves factoring and rationalizing to find x = 3 Β± 5β2/2. The third example demonstrates solving a more complex equation, resulting in x = 5/6 Β± β145/6. Each step is clearly explained, ensuring a comprehensive understanding of the method.
Takeaways
- π The lesson is about solving quadratic equations by completing the square.
- π To complete the square, take half of the coefficient of 'x', square it, and add it to both sides of the equation.
- βοΈ Ensure balance by performing the same operation on both sides of the equation.
- π Factor the quadratic expression after adding the square term to get a product of binomials.
- π’ Apply the square root to both sides to solve for 'x', resulting in two potential solutions.
- π For the equation \( x^2 + 4x = 5 \), the solutions are \( x = 1 \) and \( x = -5 \).
- π Demonstrated solving a quadratic equation by completing the square with a leading coefficient other than 1.
- π Factored out the leading coefficient and completed the square for the equation \( 2x^2 - 12x - 7 = 0 \).
- π Found the solutions for the second example to be \( x = 3 + \frac{5\sqrt{2}}{2} \) and \( x = 3 - \frac{5\sqrt{2}}{2} \).
- π Showed the process of solving a quadratic equation with a non-integer coefficient by completing the square.
- π The third example equation \( 3x^2 - 5x = 10 \) was solved, resulting in \( x = \frac{5}{6} \pm \frac{\sqrt{145}}{6} \).
Q & A
What is the method used in the script to solve quadratic equations?
-The script uses the method of completing the square to solve quadratic equations.
What is the first step in completing the square for the equation x squared plus 4x equals 5?
-The first step is to take half of the coefficient of x, which is 2 (half of 4), and square it to get 4, then add this to both sides of the equation.
How is the expression x squared plus 4x plus 4 factored in the script?
-The expression is factored as (x + 2)(x + 2), which is the product of (x + the number that was squared).
What are the solutions for x in the first example after completing the square and taking the square root of both sides?
-The solutions for x are x = 1 and x = -5, obtained by subtracting 2 from both sides of the equations x + 2 = 3 and x + 2 = -3, respectively.
How does the script handle the second example with the equation 2x squared minus 12x minus 7 equals zero?
-The script first moves the constant term to the other side, factors out the 2, completes the square by adding 18 to both sides, and then solves for x by taking the square root of both sides.
What is the process to rationalize the denominator in the second example's solution?
-The process involves multiplying the numerator and denominator by the square root of 2 to eliminate the square root in the denominator.
What is the final simplified form of the solutions for the second example in the script?
-The final simplified form of the solutions is x = 3 plus or minus 5 root 2 over 2.
How does the script approach the third example with the equation three x squared minus five x equals ten?
-The script factors out a 3 from the first two terms, completes the square by adding 25/36 to both sides, and then solves for x by taking the square root of both sides.
What is the common mistake to avoid when completing the square, as implied in the script?
-The common mistake to avoid is not squaring the entire term that is added to both sides of the equation.
Why is it necessary to multiply both sides by 1/3 in the third example in the script?
-It is necessary to eliminate the coefficient of x squared (which is 3) by multiplying both sides by 1/3, simplifying the equation to a form where the square root can be taken.
What are the final solutions for x in the third example after completing the square and simplifying?
-The final solutions for x are x = 5/6 plus or minus root 145/6.
Outlines
π Solving Quadratic Equations by Completing the Square
This paragraph introduces the method of solving quadratic equations by completing the square. The example given is \( x^2 + 4x = 5 \). The process involves taking half of the coefficient of x, squaring it, and adding it to both sides of the equation. The equation is then factored into \( (x + 2)^2 = 9 \), and the square root is taken to find two solutions: \( x = 1 \) and \( x = -5 \). The paragraph also includes a second example, \( 2x^2 - 12x - 7 = 0 \), which is solved similarly by factoring out the 2, completing the square, and finding the solutions \( x = 3 + \frac{5\sqrt{2}}{2} \) and \( x = 3 - \frac{5\sqrt{2}}{2} \).
π Completing the Square for Complex Quadratics
The second paragraph continues the theme of completing the square but with a more complex example, \( 3x^2 - 5x = 10 \). The solution starts by factoring out a 3, then taking half of the new coefficient of x and squaring it to complete the square. The equation is transformed into \( \frac{(x - \frac{5}{6})^2}{3} = \frac{145}{12} \). The square root of both sides is taken, leading to the final solutions \( x = \frac{5}{6} + \frac{\sqrt{145}}{6} \) and \( x = \frac{5}{6} - \frac{\sqrt{145}}{6} \). The paragraph also discusses the challenges of simplifying the square root of 145 and presents the final answer as a single fraction with a common denominator.
Mindmap
Keywords
π‘Quadratic Equations
π‘Completing the Square
π‘Factoring
π‘Square Root
π‘Coefficient
π‘Linear Term
π‘Perfect Square Trinomial
π‘Rationalizing the Denominator
π‘Common Denominator
π‘Solving for Variable
Highlights
Introduction to solving quadratic equations by completing the square.
Demonstration of completing the square for the equation x^2 + 4x = 5.
Explanation of adding 2^2 to both sides of the equation to balance it.
Factoring the quadratic expression x^2 + 4x + 4 into (x + 2)(x + 2).
Calculating the right side of the equation as 5 + 4 = 9.
Solving for x by taking the square root of both sides, resulting in x + 2 = Β±3.
Finding the solutions x = 1 and x = -5 by isolating x.
Starting the second example with the equation 2x^2 - 12x - 7 = 0.
Factoring out a 2 from the equation and completing the square for x^2 - 6x.
Adding 18 to both sides to balance the equation after completing the square.
Factoring the new quadratic expression x^2 - 6x + 9 into (x - 3)(x - 3).
Solving the equation by taking the square root of both sides, resulting in x - 3 = Β±β(25/2).
Rationalizing the denominator and finding the solutions x = 3 Β± (5β2)/2.
Starting the third example with the equation 3x^2 - 5x = 10.
Factoring out a 3 and completing the square for x^2 - (5/3)x.
Adding (5/6)^2 * 3 to both sides to balance the equation.
Factoring the new quadratic expression and finding the right side of the equation.
Solving for x by taking the square root of both sides and simplifying the equation.
Final solutions for x expressed as x = 5/6 Β± β145/6.
Transcripts
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