How do you find the roots, real and imaginary, of #y= 8x^2 - 10x + 14-(3x-1)^2 # using the quadratic formula?
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To find the roots of ( y = 8x^2 - 10x + 14 - (3x - 1)^2 ) using the quadratic formula, we first need to rewrite the equation in the form ( ax^2 + bx + c = 0 ). Then, we can identify ( a ), ( b ), and ( c ) to use in the quadratic formula ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).
Expanding ( (3x - 1)^2 ), we get ( 9x^2 - 6x + 1 ). Now, our equation becomes ( y = 8x^2 - 10x + 14 - (9x^2 - 6x + 1) ).
Combining like terms, we get ( y = 8x^2 - 10x + 14 - 9x^2 + 6x - 1 = -x^2 - 4x + 13 ).
Now, we have ( a = -1 ), ( b = -4 ), and ( c = 13 ). Substituting these values into the quadratic formula:
[ x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(-1)(13)}}}}{{2(-1)}} ]
[ x = \frac{{4 \pm \sqrt{{16 + 52}}}}{{-2}} ]
[ x = \frac{{4 \pm \sqrt{{68}}}}{{-2}} ]
[ x = \frac{{4 \pm 2\sqrt{{17}}i}}{{-2}} ]
[ x = -2 \pm \sqrt{{17}}i ]
Therefore, the roots of the given equation are ( -2 + \sqrt{17}i ) and ( -2 - \sqrt{17}i ), where ( i ) is the imaginary unit.
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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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