What are all the roots of the equation x^2+6x-9

Mathematics

1 answer:

frez [133]3 years ago

5 0

Answer:

x = - 3 ± 3 $\sqrt{2}$

Step-by-step explanation:

given the equation x² + 6x - 9 = 0

We can solve for x using the quadratic formula

x = ( - b ± $\sqrt{b^2-4ac }$ ) / 2a

with a = 1, b = 6 and c = - 9

x = ( - 6 ± $\sqrt{6^2-(4(1)(-9)}$ ) / 2

= ( - 6 ± $\sqrt{36+36}$ ) / 2

= ( - 6 ± $\sqrt{72}$ ) / 2

= ( - 6 ± 6 $\sqrt{2}$ ) / 2

x = - 3 ± 3 $\sqrt{2}$

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Find the area of the shape!!

Evgen [1.6K]

9514 1404 393

Answer:

100(π+3) ≈ 614 square units

Step-by-step explanation:

Assuming the rectangles are squares, the total area is that of 3 squares and a circle of radius 10.

circle area = πr² = π·10² = 100π

area of squares = 3×s² = 3·10² = 300

Then the total area is ...

300 +100π ≈ 614 . . . square units

6 0

3 years ago

Kathy lives directly east of the park. The football field is directly south of the park. The library sits on the line formed bet

enyata [817]

Look at the picture in the attachment.

Using the Pythagorean theorem, set up a system of three equations:
$x^2+y^2=(12+9)^2 \\
12^2+z^2=x^2 \\
9^2+z^2=y^2 \\ \\
x^2+y^2=441 \\
144+z^2=x^2 \\
81+z^2=y^2$

$\hbox{substitute } 144+z^2 \hbox{ for } x^2 \hbox{ and } 81+z^2 \hbox{ for } y^2 \hbox{ in the first equation:} \\
144+z^2+81+z^2=441 \\
225+2z^2=441 \\
2z^2=441-225 \\
2z^2=216 \\
z^2=\frac{216}{2} \\
z^2=108 \\
z=\sqrt{108} \\
z=\sqrt{36 \times 3} \\
z=6\sqrt{3} \\ z \approx 10.39$

$81+z^2=y^2 \\
81+108=y^2 \\
189=y^2 \\
\sqrt{189}=y \\
\sqrt{9 \times 21}=y \\
y=3\sqrt{21} \\ y \approx 13.75$

a. The library is approximately 10.39 miles (exactly: 6√3 miles) from the park.
b. The park is approximately 13.75 miles (exactly: 3√21 miles) from the football field.