Solve each quadratic equation. If roots are not real, use ii.(3z−1)2+5=0\left(3z-1\right)^2+5=0

z = 1 ± 5 i 3 z=\frac{1\pm\sqrt5i}{3} z = 3 1 ± 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> i

Solve each quadratic equation. If roots are not real, use i i . ( 3 z − 1 ) 2 + 5 = 0 \left(3z-1\right)^2+5=0

Here, we're going to solve this quadratic equation using our square root property. So we want to isolate our variable z here in order to find what value of z makes this equation true. So to start we're going to subtract five from both sides so that we get three z minus one all squared equals negative five. Now that we have our quantity squared equal to some value, we can apply our square root property. So we're going to square root both sides. So our square root and our square cancel out, and we're left with three z minus one is equal to positive and negative square root of negative five. Next, we'll add one to both sides of the equation. So we have a three z is equal to one plus or minus the square root of negative five. And next, let's divide both sides by three. So that we get z is equal to one plus or minus the square root of negative five all over three. So now we want to simplify our radical. And I know through the product property that I can think of this as the square root of five times the square root of negative one. And we know that the square root of negative one is I and that the square root of five is prime, so it can't be simplified any further. So that means our final answer is z is equal to one plus or minus the square root of five times I all over three. And if we wanted to write these as two separate values, we could also state our answer as z is equal to one plus the square root of five times I all over three. And z is equal to one minuteus the square root of five times I all over three. Both of these would be acceptable answers for our solution set.

Solve each quadratic equation. If roots are not real, use i i . ( 3 z − 1 ) 2 + 5 = 0 \left(3z-1\right)^2+5=0

z = 1 ± 5 i 3 z=\frac{1\pm\sqrt5i}{3} z = 3 1 ± 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> i

z = 1 ± 2 i 3 z=\frac{1\pm2i}{3} z = 3 1 ± 2 i

z = ± 1 3 i z=\pm\frac13i z = ± 3 1 i

10. Quadratic Equations

Complex Solutions of Quadratic Equations

Beginning Algebra

10. Quadratic Equations

Complex Solutions of Quadratic Equations

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Multiple Choice

Solve each quadratic equation. If roots are not real, use $i$ .
${(3 z - 1)}^{2} + 5 = 0$

$z=\frac{1\pm2i}{3}$

$z=\frac{1\pm\sqrt5i}{3}$

$z=\pm3i$

$z=\pm\frac13i$

Verified step by step guidance

Start with the given equation: $\left(3z - 1\right)^2 + 5 = 0$.

Isolate the squared term by subtracting 5 from both sides: $\left(3z - 1\right)^2 = -5$.

Take the square root of both sides, remembering to include the plus-minus symbol and the imaginary unit $i$ because the right side is negative: \$3z - 1 = \pm \sqrt{-5} = \pm \sqrt{5}i$.

Solve for $z$ by adding 1 to both sides: \$3z = 1 \pm \sqrt{5}i$.

Finally, divide both sides by 3 to isolate $z$: $z = \frac{1 \pm \sqrt{5}i}{3}$.

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