How do you find the roots, real and imaginary, of #y= 8x^2-15x-4-2(3x+3)^2 # using the quadratic formula?

Answer 1

#x~~-4.83# or #x~~-0.27#

First multiply out the terms so that the expression can be rearranged into standard form. #y = 8x^2 - 15x - 4 -2(9x^2+ 18x +9)# #y = 8x^2 - 18x^2 -15x -36x - 4 - 9# #y = -10x^2-51x -13# For a quadratic expression #y=ax^2 + bx+c# the quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a# #x = (-(-51)+-sqrt((-51)^2 -4(-10)(-13)))/(2(-10)# #x=(51+-sqrt(2601-520))/-20# #x = -(51+-sqrt(2081))/20# Because the number under the square root sign is positive, there are no imaginary roots, only real ones. #x~~-(51+-45.62)/20# #x~~-4.83# or #x~~-0.27#
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Answer 2

To find the roots of ( y = 8x^2 - 15x - 4 - 2(3x + 3)^2 ) using the quadratic formula, first, express the equation in the standard form ( ax^2 + bx + c = 0 ). Then, identify the values of ( a ), ( b ), and ( c ), and substitute them into the quadratic formula ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ). Finally, calculate the roots.

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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.

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.

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.

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.

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