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How do you solve # 9a=-a²-18#?

Answer 1

#a=-3#, and #a=-6#.

Start bringing everything on the same side changing the sing when you "jump" the #=# symbol.

#a^2+9a+18=0#

Now you can use the formula for the second order equation

#a=(-9\pmsqrt(81-4*18))/2=(-9\pmsqrt(9))/2=(-9\pm3)/2#. The solution with the #+# is #a=-3# and the solution with the #-# is #a=-6#.

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Answer 2

Jace Anderson

To solve the equation 9a = -a² - 18, follow these steps:

Move all terms to one side of the equation to set it equal to zero: a² + 9a - 18 = 0
Try to factor the quadratic equation. If factoring is not possible, use the quadratic formula: a = (-b ± √(b² - 4ac)) / (2a)

where a = 1, b = 9, and c = -18.
Substitute the values into the quadratic formula: a = (-(9) ± √((9)² - 4(1)(-18))) / (2(1))
Simplify the expression inside the square root: a = (-9 ± √(81 + 72)) / 2
Further simplify: a = (-9 ± √153) / 2
The solutions are the values of 'a' obtained by replacing the ± with both the positive and negative options: a₁ = (-9 + √153) / 2 a₂ = (-9 - √153) / 2

These are the solutions to the equation 9a = -a² - 18.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.