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

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Answer 1 · 2017-04-12T10:46:38

There are no imaginary roots as the graph has x-intercepts.

$" "x~~-4.45 and x~~0.45$ to 2 decimal places

First we need to change the given equation so that it is in the form $y=ax^2+bx+c$

Square the brackets

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

$y=3x^2-2x^2+4x-2$

$y=x^2+4x-2$

To determine the roots set $y=0$

$0=x^2+4x-2$

So we have: $a=1"; "b=4"; "c=-2$

$x=(-b+-sqrt(b^2-4ac))/(2a)" "->x=(-4+-sqrt((-4)^2-4(1)(-2)))/(2(1))$

$" "x=-2+-sqrt(16+8)/2$

$" "x=-2+-sqrt(24)/2$

But $24=2xx3xx4$ so $sqrt(24)=2sqrt(6)$

$" "x=-2+-(2sqrt(6))/2$

$" "x=-2+-sqrt(6)$

$" "x~~-4.45 and x~~0.45$
Tony B

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