Question

Solve the equation 2, x, squared, minus, 19, x, plus, 2, equals, minus, 10, x2x
2
−19x+2=−10x to the nearest tenth.

Answer 1

To solve the equation \( 2x^2 - 19x + 2 = -10x \), we first rearrange it into standard form:

\[
2x^2 - 19x + 2 + 10x = 0
\]

This simplifies to:

\[
2x^2 - 9x + 2 = 0
\]

Next, we can use the quadratic formula to solve for \( x \). The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation \( 2x^2 - 9x + 2 = 0 \), the coefficients are:

- \( a = 2 \)
- \( b = -9 \)
- \( c = 2 \)

Now we can calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-9)^2 - 4 \times 2 \times 2 = 81 - 16 = 65
\]

Now we can substitute into the quadratic formula:

\[
x = \frac{-(-9) \pm \sqrt{65}}{2 \times 2} = \frac{9 \pm \sqrt{65}}{4}
\]

Calculating the two possible values for \( x \):

1. \( x_1 = \frac{9 + \sqrt{65}}{4} \)
2. \( x_2 = \frac{9 - \sqrt{65}}{4} \)

Now, let's compute \( \sqrt{65} \):

\[
\sqrt{65} \approx 8.0623
\]

Using this to find the values of \( x \):

1. \( x_1 \approx \frac{9 + 8.0623}{4} = \frac{17.0623}{4} \approx 4.2656 \)
2. \( x_2 \approx \frac{9 - 8.0623}{4} = \frac{0.9377}{4} \approx 0.2344 \)

To the nearest tenth, the solutions are:

- \( x_1 \approx 4.3 \)
- \( x_2 \approx 0.2 \)

Thus, the two solutions to the equation \( 2x^2 - 19x + 2 = -10x \) are \( x \approx 4.3 \) and \( x \approx 0.2 \).