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The Quadratic Formula is a powerful tool that helps to find the real solutions to a quadratic equation, but that is not all. It is also useful for finding any complex solutions of quadratic equations. Other factoring methods can also be used to factor quadratic equations with complex roots.
### Catch-Up and Review

**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 $h=6t−t_{2}.$

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.

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After playing with her friends, Jordan went home to do her homework. She had to find the solutions to the quadratic equation $x_{2}−2x+2=0.$ Jordan knows that the Quadratic Formula is useful for finding the rational and irrational solutions to a quadratic equation. Therefore, she made a plan to use it.
When Jordan saw the negative number under the square root, she thought she had made a mistake and checked her calculations. However, everything was correct. For a moment she did not know what else to do, so she went to Mark's apartment. To help her, Mark rewrote the radical in terms of the imaginary unit and continued with the computations.
Mark told Jordan that the solutions to the equation are $x=1+i$ and $x=1−i.$ After thanking him, Jordan realized that the Quadratic Formula is even more powerful than she thought and made a very clever claim.

$Quadratic Formula x=2a-b±b_{2}−4ac $

She started by identifying the values of $a,$ $b,$ and $c$ for the given equation. After comparing her equation with the standard form of a quadratic equation $ax_{2}+bx+c=0,$ she determined that $a=1,$ $b=-2,$ and $c=2.$ Next, she substituted these values into the Quadratic Formula.
$x=2a-b±b_{2}−4ac $

Substitute values and evaluate

SubstituteValues

Substitute values

$x=2(1)-(-2)±(-2)_{2}−4(1)(2) $

CalcPowProd

Calculate power and product

$x=2-(-2)±4−8 $

NegNeg

$-(-a)=a$

$x=22±4−8 $

SubTerm

Subtract term

$x=22±-4 $

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.

She discovered that if a quadratic equation with real coefficients has one complex solution, the complex conjugate of that solution is another solution as well. In other words, if $z=a+bi$ is a solution to a quadratic equation, then $zˉ=a−bi$ is also a solution. After this, Jordan checked the solutions Mark found and verified that they meet this condition.
Jordan's teacher loves doing pendulum jumps off bridges and canyons. The teacher showed her students a video clip of how the jumper's trajectory repeats a path similar to a parabola after the rope is stretched. According to the teacher, her last jump can be modeled by the quadratic equation $h=s_{2}−8s+29,$ with $0≤s≤8.$
Here, $h$ is the distance, in meters, from the lake to the teacher $s$ seconds after the rope is played out completely and becomes taut.
### Hint

### Solution

Since the discriminant is negative, the equation has no real solutions. In this context, it means that the jumper never reached $10$ meters above the lake after the rope became taut. In other words, the jumper was always more than $10$ meters above the lake.
The solutions to the equation are $s_{1}=4−3 i$ and $s_{2}=4+3 i.$

External credits: Pseudopanax

a After the rope becomes taut, how many seconds did it take for the teacher to be $10$ meters above the lake?

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c Jordan's teacher showed a second video clip that shows a jump described by the equation $x_{2}−10x+K=0,$ where $K$ is a real number. The teacher informs the students that $x_{1}=5−2i$ is one solution to the equation. What is the other solution?

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a Substitute $10$ for $h$ in the equation modeling the jump. Before solving the equation, study the discriminant.

b Use the Quadratic Formula to solve the equation. If necessary, use complex numbers.

c If a number $a+bi$ is a solution to a quadratic equation with real coefficients, its conjugate is also a solution. To find the conjugate, change the sign of the imaginary part.

a The distance between the jumper and the lake is given by the quadratic equation $h=s_{2}−8s+29.$ Since Jordan needs to determine when the jumper was $10$ meters above the lake, substitute $10$ for $h.$

$10=s_{2}−8s+29 $

Group all the terms on the same side of the equation by subtracting $10$ from both sides.
$s_{2}−8s+19=0 $

Note that the discriminant can be evaluated before solving the equation. To do so, start by identifying the values of $a,$ $b,$ and $c.$
$s_{2}−8s+19=0⇓a=1,b=-8,c=19 $

Next, substitute these values into the expression representing the discriminant — namely, $b_{2}−4ac.$
$b_{2}−4ac$

SubstituteValues

Substitute values

$(-8)_{2}−4(1)(19)$

CalcPowProd

Calculate power and product

$64−76$

SubTerm

Subtract term

$-12$

b Although the equation set in the previous part has no real solutions, it actually does have solutions. However, they are complex numbers, so they do not make sense in a real-life context. To find them, substitute the coefficients and the discriminant found in Part A into the Quadratic Formula.

$s=2a-b±b_{2}−4ac $

Substitute values and simplify

$s=28±-12 $

SqrtNegToSqrtI

$-a =a ⋅i$

$s=28±12 i $

Simplify right-hand side

SplitIntoFactors

Split into factors

$s=28±4⋅3 i $

SqrtProd

$a⋅b =a ⋅b $

$s=28±4 ⋅3 i $

CalcRoot

Calculate root

$s=28±23 i $

FactorOut

Factor out $2$

$s=22(4±3 i) $

SimpQuot

Simplify quotient

$s=4±3 i$

c Jordan's teacher gave the class an incomplete equation and one of its solutions.

$Equation:Solution: x_{2}−10x+K=0x_{1}=5−2i $

Jordan's friends wondered how they could find the other solution without knowing the value of $K.$ Suddenly, Jordan remembered and shared with her friends the fact that she discovered last weekend.
$If a numberz=a+biis a solution to aquadratic equation with real coefficients,then its conjugatezˉ=a−biis alsoa solution. $

According to this information, the second solution is the conjugate of $x_{1}=5−2i.$ To find the conjugate of a complex number, change the sign of the imaginary part.
$x_{2}x_{2}x_{2} =x_{1} ⇓=5−2i ⇓=5+2i $

Therefore, the second solution is $x_{2}=5+2i.$
Find the solutions to the given quadratic equation. Write the answer in the form $a+bi.$
### Extra

How to Input the Answer

- If $a=0,$ omit it in the answer. That is, write the answer in the form $bi.$
- If $b=1$ or $b=-1,$ write the number in the form $a+i$ or $a−i,$ respectively.

By now, Jordan felt quite comfortable with her handling of the Quadratic Formula and wanted to know if there were more techniques for factoring equations. Then she remembered the formula for rewriting a difference of squares.

$a_{2}−b_{2}=(a+b)(a−b) $

She also remembered that at the time she originally learned that formula, she was a bit puzzled that there was no formula for the sum of squares.
$a_{2}+b_{2}=? $

She set out to explore this formula once again. Her goal was to relate the difference of squares formula to this expression. She realized that in order to do so, she needed a minus sign between the numbers, which led her to rewrite the second term.
Now she had a minus sign between the terms. However, the second term should be a single term squared. Jordan saw that in her expression, it was not. She was thinking about how to rewrite the minus sign next to $b$ as a power when the definition of the imaginary unit came to mind. She rewrote the minus sign inside the parentheses as $i_{2}.$

Jordan was excited about how things were going so far. Next, she used the Product of Powers Property to group $i$ and $b$ under the same exponent. For convenience, she wrote the imaginary unit to the right of $b.$

Finally, Jordan's expression was a difference of squares and she could apply the corresponding formula to rewrite the right-hand side expression.

By the Transitive Property of Equality — and Jordan's clever work — a formula for the sum of two squares was obtained.

$a_{2}+b_{2}=(a+bi)(a−bi)$

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.

a Maya started by opening Jordan's and Ali's solutions to the first question, in which they had to solve the equation $x_{2}+9=0.$

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b The second equation to solve is $x_{2}+7=0.$ This time, Maya opened two windows with Jordan's and Heichi's solutions.

{"type":"choice","form":{"alts":["Only Jordan","Only Heichi","Both","Neither"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}

a Use the formula $a_{2}+b_{2}=(a+bi)(a−bi).$

b Rewrite $7$ as $(7 )_{2}$ and use the formula $a_{2}+b_{2}=(a+bi)(a−bi).$

a To determine who solved the equation correctly, factor the given equation. To start, note that $9$ can be rewritten as $3_{2}.$

$x_{2}+9=0⇓x_{2}+3_{2}=0 $

The left-hand side expression is a sum of two squares and can be factored using complex numbers by using the following formula.
$a_{2}+b_{2}=(a+bi)(a−bi) $

In this case, $a=x$ and $b=3.$ Therefore, the given equation can be factored as follows.
$x_{2}+3_{2}=0⇓(x+3i)(x−3i)=0 $

From here, the solutions to the given equation are $x_{1}=3i$ and $x_{2}=-3i.$ Consequently, Ali solved the equation correctly. In contrast, Jordan made a mistake when applying the formula. She did not rewrite $9$ as $3_{2}$ and used $9$ instead of $3$ for $b.$
b As before, factor the given equation. Note that $7$ is not a perfect square and therefore cannot be rewritten as an integer squared. However, it can be written as $(7 )_{2}.$

$x_{2}+7=0⇓x_{2}+(7 )_{2}=0 $

Once again, the left-hand side expression is the sum of two squares and can be expressed as a product of two complex numbers.
$a_{2}+b_{2}=(a+bi)(a−bi) $

In this case, $a=x$ and $b=7 .$ Consequently, the given equation can be factored as follows.
$x_{2}+(7 )_{2}=0⇓(x+7 i)(x−7 i)=0 $

The solutions to the second equation are $x_{1}=7 i$ and $x_{2}=-7 i.$ Comparing the computations and solutions to the responses from the students, it can be seen that Jordan answered correctly this time. In contrast, Heichi made a mistake. Because he forgot to rewrite $7$ as $(7 )_{2},$ he took $7$ instead of $7 $ for $b.$
After hours of homework, Jordan concluded that for a quadratic equation $ax_{2}+bx+c=0,$ if $b=0,$ then the equation can be factored using either the difference of squares formula or the sum of squares formula.

However, Jordan just learned that if a quadratic equation with real coefficients has one complex solution, the conjugate of that solution is also a solution to the equation. This made her come up with a good question.

$x_{2}−y_{2}x_{2}+y_{2} =(x+y)(x−y)=(x+yi)(x−yi) $

Jordan noted that in both cases, the equation has two different solutions. She started wondering what would happen if the equation had only one solution. She remembered that in those cases, the quadratic equation has the form of a perfect square trinomial. Perfect Square Trinomial $a_{2}±2ab+b_{2}=(a±b)_{2}$ | ||
---|---|---|

Equation | Factorization | Solutions |

$x_{2}+6x+9=0$ | $(x+3)_{2}=0$ | $x_{1}=x_{2}=-3$ |

$x_{2}−8x+16=0$ | $(x−4)_{2}=0$ | $x_{1}=x_{2}=4$ |

$Can a quadratic equation haveonly one complex solution? $

For simplicity, she considered a quadratic equation that has only one pure imaginary solution. The factorization of such equation should have the form $(x−bi)_{2}=0.$ Jordan realized that expanding the square would give her a clue about how the quadratic equation looks.
After expanding the square, Jordan noticed that the quadratic equation on the right-hand side has a complex coefficient and that, in contrast to the perfect square trinomials she had studied before, the constant term is negative.

$(x±bi)_{2}=x_{2}±2bi ⋅x↑− b_{2} $

Regardless of these new facts, Jordan was very happy because she discovered a way to factor certain types of quadratic equations with complex coefficients on her own. $x_{2}±2bix−b_{2}=(x±bi)_{2}$

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.

a Dominika thought that a quadratic equation with complex coefficients would scare Jordan. For her first question, she asked Jordan to solve $x_{2}−12ix−32=4.$

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b Dominika wanted to make things even more complicated, so she decided to use complex coefficients that also involve radical numbers. The second equation set by Dominika is $3x_{2}+610 ix−30=0.$

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c As the last effort, Dominika asked for the solutions to $x_{2}+4ix−29=0.$

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a Group all the constants on the left-hand side. Rewrite the quadratic expression using the fact that $(x−bi)_{2}=x_{2}−2bix−b_{2}.$

b Divide both sides of the equation by $3.$ Note that $10$ can be rewritten as $(10 )_{2}.$ As in Part A, rewrite the quadratic expression as a perfect square trinomial with an imaginary term.

c Rewrite $-29$ as $-4−25.$ Move $-25$ to the right-hand side and factor the left-hand side expression as a perfect square trinomial with an imaginary term. Then, solve the equation for $x.$

a Since Jordan cannot use the Quadratic Formula, she began by writing down the formulas to factor a quadratic equation as a perfect square trinomial with an imaginary term.

$(x+bi)_{2}(x−bi)_{2} =x_{2}+2bix−b_{2}=x_{2}−2bix−b_{2} $

The first equation that Jordan has to solve is $x_{2}−12ix−32=4.$ Since the coefficient of the $x-$term is negative, Jordan will use the second formula. To do so, she plans to group all the terms on the same side and rewrite each of them so that they resemble the formula.
$x_{2}−12ix−32=4$

SubEqn

$LHS−4=RHS−4$

$x_{2}−12ix−36=0$

SplitIntoFactors

Split into factors

$x_{2}−2(6i)x−36=0$

Rewrite

Rewrite $36$ as $6_{2}$

$x_{2}−2(6i)x−6_{2}=0$

$x_{2}−2(6i)x−6_{2}=0⇓(x−6i)_{2}=0 $

From the resulting equation, Jordan concluded that the given equation has only one solution, $6i.$ Seeing how easily Jordan solved the first equation, Dominika began to worry.
b The first thing Jordan noticed about the second equation is that every coefficient is a multiple of $3.$ Therefore, to make her calculations easier on herself, she factored it out.

$3x_{2}+610 ix−30=0⇓3(x_{2}+210 ix−10)=0 $

Next, Jordan noted that she needs to rewrite the constant term in terms of $10 .$ She realized that $10$ can be rewritten as $(10 )_{2}.$
$3(x_{2}+210 ix−10)=0⇓3(x_{2}+210 ix−(10 )_{2})=0 $

Since the coefficient of the $x-$term is positive, Jordan will use the first formula written in the previous part.
$3(x_{2}+210 ix−(10 )_{2})=0$

DivEqn

$LHS/3=RHS/3$

$x_{2}+210 ix−(10 )_{2}=0$

$x_{2}+2bix−b_{2}=(x+bi)_{2}$

$(x+10 i)_{2}=0$

c The third equation written by Dominika does not seem like it can be factored as a perfect square trinomial. A little flustered, Jordan began by rewriting the coefficient of the $x-$term.

$x_{2}+4ix−29=0⇓x_{2}+2(2i)x−29=0 $

After this, Jordan noticed that the linear term is $2(2i)x$ implying that $b=2i.$ Consequently, $b_{2}=-4.$ This led her to conclude that she needs to have $-4$ as a constant term somehow. Therefore, she decided to rewrite $-29$ as $-4−25.$
$x_{2}+2(2i)x−29=0⇓x_{2}+2(2i)x−4−25=0 $

The first three terms can be rewritten as a perfect square trinomial with an imaginary term. Then, the resulting equation can be solved for $x.$
$x_{2}+2(2i)x−4−25=0$

AddEqn

$LHS+25=RHS+25$

$x_{2}+2(2i)x−4=25$

Rewrite

Rewrite $4$ as $2_{2}$

$x_{2}+2(2i)x−2_{2}=25$

$x_{2}+2bix−b_{2}=(x+bi)_{2}$

$(x+2i)_{2}=25$

SqrtEqn

$LHS =RHS $

$x+2i=±5$

SubEqn

$LHS−2i=RHS−2i$

$x=±5−2i$

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 $h=6t−t_{2}.$

The goal is to determine whether Mark would be able to see the ball. In other words, it should be checked whether the ball can reach a height of at least $10$ meters. To find this information, substitute $10$ for $h$ into the function rule that describes the path of the ball.$10=6t−t_{2}⇕t_{2}−6t+10=0 $

Next, investigate whether or not the resulting equation has real solutions by calculating the discriminant. To do so, use $a=1,$ $b=-6,$ and $c=10.$
$b_{2}−4ac$

Substitute values and simplify

SubstituteValues

Substitute values

$(-6)_{2}−4(1)(10)$

CalcPowProd

Calculate power and product

$36−40$

SubTerm

Subtract term

$-4$

It can also be determined graphically if the equation $10=6t−t_{2}$ has real solutions. To do so, graph $y=10$ and $y=6t−t_{2}$ on the same coordinate plane.

As can be seen, the graphs do not intersect, which means that the equation has no real solutions. However, the equation$10=6t−t_{2}$

AddEqn

$LHS+t_{2}−6t=RHS+t_{2}−6t$

$t_{2}−6t+10=0$

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a=1,b=-6,c=10$

$t=2(1)-(-6)±(-6)_{2}−4(1)(10) $

CalcPowProd

Calculate power and product

$t=2-(-6)±36−40 $

SubTerm

Subtract term

$t=2-(-6)±-4 $

NegNeg

$-(-a)=a$

$t=26±-4 $

SqrtNegToSqrtI

$-a =a ⋅i$

$t=26±4 i $

CalcRoot

Calculate root

$t=26±2i $

FactorOut

Factor out $2$

$t=22(3±i) $

ReduceFrac

$ba =b/2a/2 $

$t=3±i$