Which are the solutions to the quadratic equation x^2=9x+6

1 answer:

5 0

I will use a quadratic formula:

$x^2=9x+6\ \ \ \ |-9x-6\\\\x^2-9x-6=0\\\\a=1;\ b=-9;\ c=-6\\\\b^2-4ac=(-9)^2-4\cdot1\cdot(-6)=81+24=105\\\\\sqrt{b^2-4ac}=\sqrt{105}\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-(-9)-\sqrt{105}}{2\cdot1}=\dfrac{9-\sqrt{105}}{2}\\\\x_2=\dfrac{-(-9)+\sqrt{105}}{2\cdot1}=\dfrac{9+\sqrt{105}}{2}$

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Can you solve this? I need help. I don't understand.

ZanzabumX [31]

Answer:

2 sqrt(5) OR 4.5

Step-by-step explanation:

You have to know Pythagorean theorem to solve this question.

a^2 + b^2 = c^2

To use this theorem you have to have a right triangle. There are two right triangles in your image. The lower (larger) one has two sides labeled, so you can use Pythagorean thm to find the third side. There's a short cut, bc some right triangles have easy-to-memorize lengths of the sides. 3-4-5 is one of these number sets. A multiple of this is 6-8-10. We could've solved:

b^2 + 8^2 = 10^2

But it would've come out the same. The unlabeled side is 6.

We can use the 6 and the 4 on the smaller right triangle and use the Pythagorean thm again to solve for x.

4^2 + x^2 = 6^2

16 + x^2 = 36 subtract 16 from both sides.

x^2 = 20

Take the square root of both sides.

sqrt (x^2) = sqrt 20

x = 2 sqrt(5) which is approximately 4.472.

2 sqrt(5) is an exact answer if that is what they are asking for. 4.472 is an approximation to the nearest thousandth. It would be 4.47 to the nearest hundredth or 4.5 to the nearest tenth.

8 0

2 years ago

Find the value of the variable.

anzhelika [568]

Answer:

hope this helps you look it once.