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How do you find the roots, real and imaginary, of $y=(x-2)(x+1)-3x$ using the quadratic formula?

2 Answers

Apr 18, 2016

$y=(x-2)(x+1)-3x$
$color(white)("XXX")$ has two real roots at $x=2+sqrt(6) and x=2-sqrt(6)$

Explanation:

In order to use the quadratic formula it is first necessary to convert the given equation into standard form.

$y=(x-2)(x+1)-3x$

$rarr y=x^2-x-2-3x$

$rarr y=x^2-4x-2$ (standard form)

Roots for an equation of the form:
$color(white)("XXX")y=ax^2+bx+c$
are given by the quadratic formula
$color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case
$color(white)("XXX")a=1,$
$color(white)("XXX")b=-4$
$color(white)("XXX")c=-2$

So the roots are
$color(white)("XXX")x=(4+-sqrt((-4)^2-4(1)(-2)))/(2(1))$

$color(white)("XXX")x=(4+-sqrt(24))/2$

$color(white)("XXX")x=2+-sqrt(6)$

Shwetank Mauria

Apr 18, 2016

We have only real roots, which are $2-sqrt6$ and $2+sqrt6$

Explanation:

$y=(x-2)(x+1)-3x=x^2-2x+x-2-3x=x^2-4x-2$

The roots of $x^2-4x-2$

given by quadratic formula $(-b+-sqrt(b^2-4ac))/(2a)$ are

$x=(-(-4)+-sqrt((-4)^2-4xx1xx(-2)))/(2xx1)=(4+-sqrt(16+8))/2$ or

$x=(4+-sqrt24)/2=(4+-2sqrt6)/2=2+-sqrt6$

Hence, roots are $2-sqrt6$ and $2+sqrt6$

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