How do you find the roots, real and imaginary, of $y = - 3 {(x - 1)}^{2} - 3 x - 1$ $y=-3(x -1 )^2-3x-1$ using the quadratic formula?

Question

How do you find the roots, real and imaginary, of $y = - 3 {(x - 1)}^{2} - 3 x - 1$ $y=-3(x -1 )^2-3x-1$ using the quadratic formula?

1 Answer

Impact of this question

1526 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T17:08:33

The roots are $x = \frac{1}{2} + i \frac{\sqrt{39}}{6}$ $x=1/2+isqrt39/6$ and $x = \frac{1}{2} - i \frac{\sqrt{39}}{6}$ $x=1/2-isqrt39/6$ .

Explanation:

First, express $y$ $y$ in standard form.

$y = - 3 {(x - 1)}^{2} - 3 x - 1 = - 3 x^{2} + 6 x - 3 - 3 x - 1 = - 3 x^{2} + 3 x - 4$ $y=color(red)(-3(x-1)^2)-3x-1=color(red)(-3x^2+6x-3)-3x-1=-3x^2+3x-4$

The quadratic formula finds when $y = 0$ $y=0$ . $y = 0$ $y=0$ when

$x = \frac{- 3 \pm \sqrt{3^{2} - 4 (- 4) (- 3)}}{2 \cdot (- 3)}$ $x=(-3+-sqrt(3^2-4(-4)(-3)))/(2*(-3))$

$= \frac{- 3 \pm \sqrt{9 - 48}}{2 \cdot (- 3)} = \frac{- 3 \pm \sqrt{- 39}}{2 \cdot (- 3)} = \frac{- 3 \pm i \sqrt{39}}{2 \cdot (- 3)}$ $=(-3+-sqrt(9-48))/(2*(-3))=(-3+-sqrt(-39))/(2*(-3))=(-3+-isqrt(39))/(2*(-3))$

$= \frac{1}{2} \pm i \frac{\sqrt{39}}{6}$ $=1/2+-isqrt(39)/6$

Therefore the roots are

$x = \frac{1}{2} + i \frac{\sqrt{39}}{6}$ $x=1/2+isqrt39/6$ and $x = \frac{1}{2} - i \frac{\sqrt{39}}{6}$ $x=1/2-isqrt39/6$ .

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of $y = - 3 {(x - 1)}^{2} - 3 x - 1$ $y=-3(x -1 )^2-3x-1$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=-3(x -1 )^2-3x-1y=−3(x−1)2−3x−1y=-3(x -1 )^2-3x-1 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the roots, real and imaginary, of $y = - 3 {(x - 1)}^{2} - 3 x - 1$ $y=-3(x -1 )^2-3x-1$ using the quadratic formula?