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bry856

  • one year ago

The roots of an equation are x = -1 ± i. The equation is x^2 + _____ + 2 = 0.

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  1. sdfgsdfgs
    • one year ago
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    "The roots of an equation are x = -1 ± i. " so the eqn is (x-(-1+i))*(x-(-1-i)) = 0 expend the above n simply...what will u get?

  2. jim_thompson5910
    • one year ago
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    Let p and q be the roots of \(\Large x^2 + bx + c = 0\) Rule: \(\Large p+q = -b\) so \(\Large b = -(p+q)\)

  3. campbell_st
    • one year ago
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    well if you think about the general quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] you can see from the equation a = 1 and c = 2 so \[x = \frac{-b \pm \sqrt{b^2 - 4 \times 1 \times 2}}{2 \times 1}\] then comparing the parts \[-1 = \frac{-b}{2}\] solve for b... and then check with b by using the rest of the quadratic formula \[b^2 - 8 = \frac{\sqrt{4i^2}}{2}\]

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