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| 11 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
One afternoon, Jordan was playing soccer in front of her building with some friends. She wanted to invite her neighbor Mark over to play. To do so, she kicked the ball upward as hard as she could, trying to get Mark to see it outside his window. That kick followed the path of the quadratic equation h=6t−t2.
Here, h is the height in meters of the ball above the ground and t is the time in seconds after the kick. If Mark's bedroom window is 10 meters above the ground, could Mark see the ball? If so, find the value of t for the amount of time it takes for the ball to reach Mark's bedroom window. If not, select No real roots.
Substitute values
Calculate power and product
-(-a)=a
Subtract term
Later that evening, she was still in awe at the Quadratic Formula and wrote it on a chalkboard in her room. She then began to think about the cases in which the discriminant is negative. After analyzing the fact that there are two signs in front of the radical term, she came to another brilliant conclusion.
Substitute values
Calculate power and product
Subtract term
-a=a⋅i
Split into factors
a⋅b=a⋅b
Calculate root
Factor out 2
Simplify quotient
a2+b2=(a+bi)(a−bi)
Maya is reviewing the responses to the last two-question quiz on solving quadratic equations that her math students took through the school's online platform.
Perfect Square Trinomial a2±2ab+b2=(a±b)2 | ||
---|---|---|
Equation | Factorization | Solutions |
x2+6x+9=0 | (x+3)2=0 | x1=x2=-3 |
x2−8x+16=0 | (x−4)2=0 | x1=x2=4 |
x2±2bix−b2=(x±bi)2
Seeing Jordan's enthusiasm for solving quadratic equations, her older sister Dominika wanted to set her an interesting challenge.
Feeling confident, Jordan accepted the challenge.
LHS−4=RHS−4
Split into factors
Rewrite 36 as 62
LHS+25=RHS+25
Rewrite 4 as 22
x2+2bix−b2=(x+bi)2
LHS=RHS
LHS−2i=RHS−2i
Factor each quadratic equation.
In the challenge presented, it was said that Jordan kicked the ball up as hard as she could, trying to make it visible from Mark's bedroom window. Mark's window is 10 meters above the ground. Additionally, it is known that the ball followed the path of the quadratic equation h=6t−t2.
Substitute values
Calculate power and product
Subtract term
It can also be determined graphically if the equation 10=6t−t2 has real solutions. To do so, graph y=10 and y=6t−t2 on the same coordinate plane.
LHS+t2−6t=RHS+t2−6t
Use the Quadratic Formula: a=1,b=-6,c=10
Calculate power and product
Subtract term
-(-a)=a
-a=a⋅i
Calculate root
Factor out 2
ba=b/2a/2
Determine if the given statement is sometimes, always, or never true.
In a quadratic equation in standard form, if the signs of a and c are the same — both positive or both negative — then the solutions will be imaginary. |
Recall that the discriminant of a quadratic equation can be used to determine the number and type of solutions to the equation. lc Quadratic equation:& ax^2+ bx+ c=0 [0.3em] Solutions:&x=- b±sqrt(b^2-4 a c)/2 a [0.8em] Discriminant:& b^2-4 a c The table below summarizes the relationship between the discriminant and the number and type of solutions.
Discriminant (b^2-4ac) | Type and Number of Solutions |
---|---|
Positive | 2 real solutions |
Zero | 1 real solution |
Negative | 2 imaginary solutions |
Let's make some observations about the terms in the discriminant.
b^2-4 a c>0 whenb^2 > 4ac [0.5em] b^2-4 a c<0 whenb^2 < 4ac [0.5em] b^2-4 a c=0 whenb^2 = 4ac If a and c have the same sign, then the discriminant can be positive, negative, or zero. Therefore, it is sometimes true that if the signs of a and c are the same — both positive or both negative — then the solutions will be imaginary.
A student tried to solve a quadratic equation. Consider his work below.
Let's examine the steps that the student took to solve the quadratic equation.
To solve a quadratic equation, we first need to write all terms on one side of the equation, as the student did in Step I. After this, he had the sum of two terms 4x^2 and 5. The student then likely recalled that the sum of two perfect squares can be written as the product of two complex conjugates. a^2+b^2 ⇔ (a+bi)(a-bi) We should consider how to write each of the terms a^2=4x^2 and b^2=5 as a perfect square in order to apply this formula. 4x^2&=(2x)^2 [0.5em] 5&=(sqrt(5))^2 Note that the student did not express the number 5 as the square of a number while applying the formula Therefore, we can conclude that the expression was factored incorrectly in Step II.
In order to solve the quadratic equation, we will first write all the terms on the left-hand side and then write the terms as a sum of two perfect squares.
We will apply the formula that we expressed in Part A. To do so, we will write the sum of (2x)^2 and (sqrt(5))^2 as the product of two complex conjugates. (2x)^2+(sqrt(5))^2=0 [0.5em] ⇕ [0.5em] (2x+sqrt(5)i)(2x-sqrt(5)i)=0 We can now solve this equation by using the Zero Product Property.
We found the solution to be - sqrt(5)2i and sqrt(5)2i. Let's show the solution steps once again by correcting the student's mistake.
Let's start by writing all terms on the left-hand side of the equation by using the Properties of Equality.
Note that every coefficient is a multiple of 2. Let's factor it out! 2x^2 + 108ix - 1508 = 0 [0.15em] ⇓ [0.15em] 2(x^2 + 54ix - 754) = 0 We can simplify the equation by dividing both sides by 2. 2(x^2 + 54ix - 754) = 0 [0.15em] ⇓ [0.15em] x^2 + 54ix - 754 = 0 Note that this quadratic equation has an imaginary term. Let's recall the formulas used to factor a quadratic equation that is a perfect square trinomial with an imaginary term. (x + bi)^2 &= x^2 + 2bix - b^2 [0.25em] (x - bi)^2 &= x^2 - 2bix - b^2 Our equation does not seem like it can be factored as a perfect square trinomial. Let's rewrite the coefficient of the x-term. x^2 + 54ix - 754 = 0 ⇓ x^2 + 2(27i)x - 754 = 0 Since the x-term is 2(27i)x, b is equal to 27i. Consequently, b^2 = -729. Therefore, we need to have -729 as a constant term somehow. Let's rewrite -754 as -729 -25. x^2 + 2(27i)x - 754 = 0 ⇓ x^2 + 2(27i)x - 729 - 25 = 0 If we move the 25 to the other side of the equation, the first three terms can be rewritten as a perfect square trinomial with an imaginary term by using the first formula because the coefficient of the x-term is positive. The resulting equation can then be solved for x.
We found that the solutions to the equation are x_1 = -5-27i and x_2 = 5-27i.
We can solve quadratic equations with complex coefficients by using the Quadratic Formula. Consider the given quadratic equation. 2x^2+21ix-1413=95-87ix In the solution, we wrote all terms on the left-hand side of the equation. 2x^2 + 108ix - 1508 = 0 Then we simplified it by dividing both sides by 2, which resulted in the following equation. x^2 + 54ix - 754 = 0 We can now apply the Quadratic Formula to solve this equation. In the Quadratic Formula, b^2-4ac is the discriminant. ax^2+bx+c=0 ⇕ x=- b±sqrt(b^2-4ac)/2a Let's identify the values of a, b, and c. x^2 + 54ix - 754 = 0 ⇕ 1x^2+( 54i)x+( - 754)=0 As we can see, a= 1, b= 54i, and c= - 754. Now let's calculate the discriminant.
The discriminant is 100. We will now substitute the values of a, b, c, and the discriminant into the Quadratic Formula.
Let's write the solutions in standard form. &x_1=- 27i + 5 ⇔ x_1=5-27i &x_2=- 27i - 5 ⇔ x_2=- 5-27i The solutions of the given equation are x_1 = -5-27i and x_2 = 5-27i.