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Solve Quadratic Equations with Complex Solutions 2Solve Quadratic Equations with Complex Solutions 2
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Question 1 of 4
1. Question
Solve
`2x^2+3x+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=2`, `b=3`, and `c=9`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=2`, `b=3`, and `c=9`. `x=` `(-color(red)(3)\+-sqrt(color(red)(3)^2-4color(blue)((2))color(green)((9))))/(2color(blue)((2)))` Simplify `x=` `(-3\+-sqrt(3^2-4(2)(9)))/(4)` Simplify `3^2` `x=` `(-3\+-sqrt(9-4(2)(9)))/(4)` Multiply `-4(2)(9)`. `x=` `(-3\+-sqrt(9-72))/(4)` Subtract under the root. `x=` `(-3\+-sqrt(-63))/(4)` Separate out the `sqrt(-1)` from the root. `x=` `(-3\+-sqrt(-1)timessqrt(63))/(4)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(-3\+-itimessqrt(9)timessqrt(7))/(4)` Simplify the root using `sqrt(63) = sqrt(9)timessqrt(7)`. `x=` `(-3\+-itimes3timessqrt(7))/(4)` Simplify the root. `x=` `(-3\+-3isqrt(7))/(4)` Rearrange `3timesitimessqrt(7)` so that the `i` is in between the constant and the root. `x=` `-3/4 \+-3isqrt(7)/4` Divide both terms in the numerator by `4`. `-3/4 + 3isqrt(7)/4` and `-3/4 – 3isqrt(7)/4` Break into the two solutions. `-3/4 + 3isqrt(7)/4` and `-3/4 – 3isqrt(7)/4` -
Question 2 of 4
2. Question
Solve
`2x^2+10x+17=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=2`, `b=10`, and `c=17`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=2`, `b=10`, and `c=17`. `x=` `(-color(red)(10)\+-sqrt(color(red)(10)^2-4color(blue)((2))color(green)((17))))/(2color(blue)((2)))` Simplify `x=` `(-10\+-sqrt(10^2-4(2)(17)))/(4)` Simplify `10^2` `x=` `(-10\+-sqrt(100-4(2)(17)))/(4)` Multiply `-4(2)(17)`. `x=` `(-10\+-sqrt(100-136))/(4)` Subtract under the root. `x=` `(-10\+-sqrt(-36))/(4)` Separate out the `sqrt(-1)` from the root. `x=` `(-10\+-sqrt(-1)timessqrt(36))/(4)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(-10\+-itimes6)/(4)` Simplify the root. `x=` `(-10\+-6i)/(4)` Rearrange `itimes6` so that the `i` is in between the constant and the root. `x=` `-10/4 \+-(6i)/4` Divide both terms in the numerator by `4`. `x=` `-5/2 \+-(3i)/2` Simplify. `-5/2 +(3i)/2` and `-5/2-(3i)/2` Break into the two solutions. `-5/2 +(3i)/2` and `-5/2-(3i)/2` -
Question 3 of 4
3. Question
Solve
`3x^2+2x+11=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=3`, `b=2`, and `c=11`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=3`, `b=2`, and `c=11`. `x=` `(-color(red)(2)\+-sqrt(color(red)(2)^2-4color(blue)((3))color(green)((11))))/(2color(blue)((3)))` Simplify `x=` `(-2\+-sqrt(2^2-4(3)(11)))/(6)` Simplify `2^2` `x=` `(-2\+-sqrt(4-4(3)(11)))/(6)` Multiply `-4(3)(11)`. `x=` `(-2\+-sqrt(4-132))/(6)` Subtract under the root. `x=` `(-2\+-sqrt(-128))/(6)` Separate out the `sqrt(-1)` from the root. `x=` `(-2\+-sqrt(-1)timessqrt(128))/(6)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(-2\+-itimessqrt(64)timessqrt(2))/(6)` Simplify the root using `sqrt(128) = sqrt(64)timessqrt(2)`. `x=` `(-2\+-itimes8timessqrt(2))/(6)` Simplify the root. `x=` `(-2\+-8isqrt(2))/(6)` Rearrange `itimes8timessqrt(2)` so that the `i` is in between the constant and the root. `x=` `-2/6 \+-(8isqrt(2))/6` Divide both terms in the numerator by `6`. `-1/3 \+-(4isqrt(2))/3` Simplify. `-1/3+(4isqrt(2))/3` and `-1/3-(4isqrt(2))/3` Break into the two solutions. `-1/3+(4isqrt(2))/3` and `-1/3-(4isqrt(2))/3` -
Question 4 of 4
4. Question
Solve
`x^2-2sqrt2x+6=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=-2sqrt2`, and `c=6`.`x=` `(color(red)(-b)\+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c)))/(2color(blue)(a))` Substitute in `a=1`, `b=-2sqrt2`, and `c=6`. `x=` `(-color(red)(-2sqrt2)\+-sqrt(color(red)(-2sqrt2)^2-4color(blue)((1))color(green)((6))))/(2color(blue)((1)))` Simplify `x=` `(-(-2sqrt2)\+-sqrt((-2sqrt2)^2-4(1)(6)))/(2)` Simplify `(-2sqrt2)^2`. `x=` `(2sqrt2\+-sqrt(8-4(1)(6)))/(2)` Multiply `-4(1)(6)`. `x=` `(2sqrt2\+-sqrt(8-24))/(2)` Subtract under the root. `x=` `(2sqrt2\+-sqrt(-16))/(2)` Separate out the `sqrt(-1)` from the root. `x=` `(2sqrt2\+-sqrt(-1)timessqrt(16))/(2)` Replace the `sqrt(-1)` remember `sqrt(-1)=i` `x=` `(2sqrt2\+-itimessqrt(16))/(2)` Simplify the root. `x=` `(2sqrt2\+-4i)/(2)` Rearrange `itimessqrt(16)` so that the `i` is on the right-hand side in this term. `x=` `(2sqrt2)/2\+-(4i)/2` Divide both terms in the numerator by `2`. `x=` `sqrt2\+-2i` `sqrt2+2i` and `sqrt2-2i` Break into the two solutions. `sqrt2+2i` and `sqrt2-2i`
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