Question

How do you find the roots, real and imaginary, of `y=4x^2 +x -3-(x-2)^2` using the quadratic formula?

Answer 1

`x=0.9067 and x=-2.5734`

Explanation:

first, expand the bracket

`(x-2)^2`

`(x-2)(x-2)`

`x^2-4x+4`

then, solve the equations

`y=4x^2+ x-3-(x^2-4x+4)`

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

`y=3x^2+ 5x-7`

then, by using `b^2-4ac`

for the equation: `y=3x^2+ 5x-7`

where `a=3, b=5 and c=-7` into `b^2-4ac`

`5^2-4(3)(-7)`

`25--84`

`109`

so, compare to this

`b^2-4ac>0` : two real and different roots
`b^2-4ac=0` : two real root and equals
`b^2-4ac<0` : no real roots or (the roots are complexes)

so, `109>0` means two real and different roots

thus, you must use this formula to find the imaginary roots

`x= (-b+- sqrt(b^2-4ac ))/ (2a)`

`x= (-5+- sqrt(5^2-4(3)(-7) ))/ (2(3)`

`x= (-5+- sqrt(109))/ 6`

`x= (-5+sqrt(109))/ 6` and `x= (-5- sqrt(109))/ 6`

solve it and u will get the values of x which is

`x=0.9067 and x=-2.5734`