Question

X ² + 13 x + 27 = 0 x 2 +13x+27=0 to the nearest tenth.

Answer 1

To solve the quadratic equation x² + 13x + 27 = 0, we use the quadratic formula with a = 1, b = 13, and c = 27, and round our answers to the nearest tenth.

Step-by-step explanation:

The equation at hand is x² + 13x + 27 = 0. This is a quadratic equation, which is an equation in the form of ax² + bx + c = 0. To find the value of x that solves this equation, we can use the quadratic formula, given by x = (-b ± √(b² - 4ac))/(2a). In our case, a = 1, b = 13, and c = 27. Substituting these values into the quadratic formula gives:

x = –(13) ± √((13)² - 4 × 1 × 27))/(2 × 1)

x = –(13) ± √(169 - 108))/(2)

x = (-13 ± √(61))/2

After calculating the square root and dividing by 2, we will get two solutions for x, and we should round the answers to the nearest tenth.

Answer 2

The quadratic equation
`x^2 + 13x + 27 = 0`, use the quadratic formula and plug in the values of a = 1, b = 13, and c = 27. The solutions to the equation are approximately x ≈ -1.2 and x ≈ -11.8.

The quadratic equation
`x^2 + 13x + 27 = 0,` you can use the quadratic formula:

`{x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}`

In your equation, a = 1, b = 13, and c = 27. Plugging these values into the formula:

So, the solutions for x are:

`x = (-13 + √(61) ) / 2`

x ≈ -1.2

`x = (-13 - √(61) ) / 2`

x ≈ -11.8

Therefore, the solutions to the quadratic equation
`x^2 + 13x + 27 = 0` to the nearest tenth are approximately x ≈ -1.2 and x ≈ -11.8.