Question

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

x^2 - 23 = 10x

a. -1.93, 11.93
b. 1.93, -11.93
c. 1.93, 11.93
d. -1.93, -11.93

Answer 1

Answer:

Option B. 1.93, -11.93 is the right answer.

Step-by-step explanation:

Here the given equation is x²-23 = 10x and we have to solve the equation for the value of x.

First we simplify the equation to bring in the shape of ax²+bx+c=0

x²-23 = 10x

(x²-23)-10x = 10x-10x

x²-10x -23 = 0

Now we the formula for the value of $x= \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{10\pm \sqrt{(-10)^{2}-4(1)(-23)}}{2(1)}$

$x=\frac{-10\pm  \sqrt{100+92}}{2}$

$x= \frac{-10\pm \sqrt{192}}{2}$

$x=\frac{-10\pm 4\sqrt{12}}{2}$

$x=-5\pm 2\sqrt{12}$ $= -5\pm (2\times 3.464)$ $= -5\pm 6.93$

Now x = (-5+6.93) = 1.93

and x = (-5-6.93) = -11.93

Answer 2

Answer:

The correct answer option is a. -1.93, 11.93.

Step-by-step explanation:

We are given the following equation and we are to solve it using the quadratic formula:

$x^2 - 23 = 10x$

Re-arranging this equation in order of decreasing power:

$x^{2} - 10x - 23 = 0$

Using the quadratic formula:

$x = \frac {-b + - \sqrt{b^2 - 4ac} }{2a}$

Substituting the given values in the formula to get:

$x=\frac{-(-10)+-\sqrt{(-10)^2-4(1)(-23)} }{2(1)}$

$x=\frac{10+-\sqrt{192} }{2}$

$x=\frac{10+\sqrt{192} }{2} , x=\frac{10-\sqrt{192} }{2}$

$x=11.93, x=-1.93$