How do you find the roots, real and imaginary, of $y= -x^2 -x+(2x- 1 )^2$ using the quadratic formula?

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Answer 1 · 2016-06-04T21:44:44

1.435 and .232

first you want to simplify the equation by expanding the brackets and then collecting like terms.

$-x^2-x+(2x-1)^2$

$-x^2-x+(2x-1)(2x-1)$

$-x^2-x+(4x^2-2x-2x+1)$

$-x^2-x+4x^2-4x+1$

$3x^2-5x+1$

$:. a=3, b=-5 and c=1$

sub values into quadratic equation

$(-b +- sqrt (b^2 - 4ac))/(2a)$

$(-(-5) +- sqrt (-5^2 - 4(3)(1)))/(2(3))$

$(5 +- sqrt (25-12))/(6)$

$(5 +- sqrt (13))/(6)$

$(5+3.61)/(6)$ and $(5-3.61)/(6)$

$1.435$ and $.232$

How do you find the roots, real and imaginary, of y= -x^2 -x+(2x- 1 )^2 y= -x^2 -x+(2x- 1 )^2 using the quadratic formula?