Topics
>
Precalculus>
Complex Numbers>
Solve Quadratic Equations with Complex Solutions>
Solve Quadratic Equations with Complex Solutions 1Solve Quadratic Equations with Complex Solutions 1
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- Answered
- Review
-
Question 1 of 4
1. Question
Solve
`x^2+6x+13=0`
Correct
Great Work!
Incorrect
To find the roots of an equation in the form `color(blue)(a)x^2+color(red)(b)x+color(green)(c)=0` where `a\ne0`, use the quadratic formula `x=(-b\+-sqrt(b^2-4ac))/(2a)`.The equation `x^2+6x+13=0` gives us `color(blue)(a=1)`, `color(red)(b=6)`, and `color(green)(c=13)`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=1`, `b=6`, and `c=13`. `x=` `(-color(red)(6)\+-sqrt(color(red)(6)^2-4color(blue)((1))color(green)((13))))/(2color(blue)((1)))` Simplify `x=` `(-6\+-sqrt(6^2-4(1)(13)))/(2)` Simplify `6^2` `x=` `(-6\+-sqrt(36-4(1)(13)))/(2)` Multiply `-4(1)(13)`. `x=` `(-6\+-sqrt(36-52))/(2)` Subtract under the root. `x=` `(-6\+-sqrt(-16))/(2)` Separate out the `sqrt(-1)` from the root. `x=` `(-6\+-sqrt(-1)timessqrt(16))/(2)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(-6\+-itimessqrt(16))/(2)` Simplify the root. `x=` `(-6\+-itimes4)/(2)` Rearrange `-itimes4` so that the `i` is on the right-hand side in this term. `x=` `(-6\+-4i)/(2)` Divide both terms in the numerator by `2`. `x=` `-3\+-2i` `x=-3+2i` and `x=-3-2i` Break into the two solutions. `x=-3+2i` and `x=-3-2i` -
Question 2 of 4
2. Question
Solve
`x^2+2x+9=0`
Correct
Great Work!
Incorrect
To find the roots of an equation in the form `color(blue)(a)x^2+color(red)(b)x+color(green)(c)=0` where `a\ne0`, use the quadratic formula `x=(-b\+-sqrt(b^2-4ac))/(2a)`.The equation `x^2+6x+13=0` gives us `a=1`, `b=2`, and `c=9`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=1`, `b=2`, and `c=9`. `x=` `(-color(red)(2)\+-sqrt(color(red)(2)^2-4color(blue)((1))color(green)((9))))/(2color(blue)((1)))` Simplify `x=` `(-2\+-sqrt(2^2-4(1)(9)))/(2)` Simplify `2^2` `x=` `(-2\+-sqrt(4-4(1)(9)))/(2)` Multiply `-4(1)(9)`. `x=` `(-2\+-sqrt(4-36))/(2)` Subtract under the root. `x=` `(-2\+-sqrt(-32))/(2)` Separate out the `sqrt(-1)` from the root. `x=` `(-2\+-sqrt(-1)sqrt(32))/(2)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(-2\+-isqrt(32))/(2)` Simplify using `sqrt(32) = sqrt(16)timessqrt(2)`. `x=` `(-2\+-isqrt(16)timessqrt(2))/(2)` Simplify the root. `x=` `(-2\+-i4sqrt(2))/(2)` Rearrange `itimes4timessqrt(2)` so that the `i` is in between the constant and the root. `x=` `(-2\+-4isqrt(2))/(2)` Divide both terms in the numerator by `2`. `x=` `-1\+-2isqrt(2)` `x=-1+2isqrt(2)` and `x=-1-2isqrt(2)` Break into the two solutions. `x=-1+2isqrt(2)` and `x=-1-2isqrt(2)` -
Question 3 of 4
3. Question
Solve
`x^2+5x+25=0`
Correct
Great Work!
Incorrect
To find the roots of an equation in the form `color(blue)(a)x^2+color(red)(b)x+color(green)(c)=0` where `a\ne0`, use the quadratic formula `x=(-b\+-sqrt(b^2-4ac))/(2a)`.The equation `x^2+6x+13=0` gives us `a=1`, `b=5`, and `c=25`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=1`, `b=5`, and `c=25`. `x=` `(-color(red)(5)\+-sqrt(color(red)(5)^2-4color(blue)((1))color(green)((25))))/(2color(blue)((1)))` Simplify `x=` `(-5\+-sqrt(5^2-4(1)(25)))/(2)` Simplify `5^2` `x=` `(-5\+-sqrt(25-4(1)(25)))/(2)` Multiply `-4(1)(25)`. `x=` `(-5\+-sqrt(25-100))/(2)` Subtract under the root. `x=` `(-5\+-sqrt(-75))/(2)` Separate out the `sqrt(-1)` from the root. `x=` `(-5\+-sqrt(-1)timessqrt(75))/(2)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(-5\+-itimessqrt(25)timessqrt(3))/(2)` Simplify the root using `sqrt(75) = sqrt(25)timessqrt(3)`. `x=` `(-5\+-itimes5timessqrt(3))/(2)` Simplify the root. `x=` `(-5\+-5isqrt(3))/(2)` Rearrange `-itimes4` so that the `i` is in between the constant and the root. `x=` `-5/2\+-(5isqrt(3))/(2)` Divide both terms in the numerator by `2`. `-5/2+(5isqrt(3))/(2)` and `-5/2\-(5isqrt(3))/(2)` Break into the two solutions. `-5/2+(5isqrt(3))/(2)` and `-5/2\-(5isqrt(3))/(2)` -
Question 4 of 4
4. Question
Solve
`3x+1/x=2`
Correct
Great Work!
Incorrect
To find the roots of an equation in the form `ax^2+bx+c=0` where `a\ne0`, use the quadratic formula `x=(-b\+-sqrt(b^2-4ac))/(2a)`.Simplify the equation `3x+1/x=2` by multiplying all of the terms by `x`. This gives `3x^2+1=2x`. Now rearrange the equation into the form `ax^2+bx+c=0`. This is `3x^2-2x+1=0`The equation `3x^2-2x+1=0` gives us `a=3`, `b=-2`, and `c=1`.`x=` `(-b\+-sqrt(b^2-4ac))/(2a)` Substitute in `a=3`, `b=-2`, and `c=1`. `x=` `(-color(red)((-2))\+-sqrt(color(red)((-2))^2-4color(blue)((3))color(green)((1))))/(2color(blue)((3)))` Simplify `-color(red)((-2))` `x=` `(2\+-sqrt((-2)^2-4(3)(1)))/(2(3))` Simplify `(-2)^2` `x=` `(2\+-sqrt(4-4(3)(1)))/(2(3))` Simplify `-4(3)(1)` `x=` `(2\+-sqrt(4-12))/(2(3))` Subtract under the root. `x=` `(2\+-sqrt(-8))/(2(3))` Multiply `2(3)` `x=` `(2\+-sqrt(-1)timessqrt(8))/(6)` Separate out the `sqrt(-1)` from the root. `x=` `(2\+-itimessqrt(8))/(6)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(2\+-itimessqrt(4times2))/(6)` Simplify the root. `x=` `(2\+-itimes2sqrt(2))/(6)` Rearrange `itimes2sqrt(2)` so that the `i` is on the right-hand side in this term. `x=` `2/6\+-(2sqrt(2))/6i` Divide both terms in the numerator by `2`. `x=` `1/3\+-sqrt(2)/3i` Simplify `x=1/3+sqrt(2)/3i` and `x=1/3-sqrt(2)/3i` Break into the two solutions. `x=1/3+sqrt(2)/3i` and `x=1/3-sqrt(2)/3i`
Quizzes
- Simplify Roots of Negative Numbers 1
- Simplify Roots of Negative Numbers 2
- Powers of the Imaginary Unit 1
- Powers of the Imaginary Unit 2
- Solve Quadratic Equations with Complex Solutions 1
- Solve Quadratic Equations with Complex Solutions 2
- Equality of Complex Numbers
- Add and Subtract Complex Numbers 1
- Add and Subtract Complex Numbers 2
- Multiply Complex Numbers 1
- Multiply Complex Numbers 2
- Divide Complex Numbers
- Complex Numbers – Product of Linear Factors 1
- Complex Numbers – Product of Linear Factors 2
- Mixed Operations with Complex Numbers