Solve 2x2 − 4x − 5 = 0 by completing the square.

1 answer:

3 0

For this case we have the following polynomial:
$2x ^ 2 - 4x - 5 = 0
$
Rewriting we have:
$2x ^ 2 - 4x = 5

x ^ 2 - 2x = 5/2$
Then, completing squares we have:
$x ^ 2 - 2x + (-2/2) ^ 2 = 5/2 + (-2/2) ^ 2
$
Rewriting:
$x ^ 2 - 2x + (-1) ^ 2 = 5/2 + (-1) ^ 2

x ^ 2 - 2x + 1 = 5/2 + 1

(x-1) ^ 2 = 7/2$
$x-1 = +/- \sqrt{7/2} x = 1 +/- \sqrt{7/2}$
Then, the solutions are:
$x1 = 1 + \sqrt{7/2} x2 = 1 - \sqrt{7/2}$
Answer:
$x1 = 1 + \sqrt{7/2} 

x2 = 1 - \sqrt{7/2}$

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Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .