Question

How do you find the roots, real and imaginary, of `y= 2x^2-15x-(4x+3)^2` using the quadratic formula?

Answer 1

`x~~-1.69" and "x~~-0.38" to 2 decimal places")`

Explanation:

You need to add like terms such that you and up with equation form:`" "y=ax^2+bx+c`

To do this expand the brackets and simplify.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Expanding the brackets")`

Note that `[-(4x+3)]^2" is not the same as "-[(4x+3)^2]`
The condition we have is`" "-[(4x+3)^2]`

`color(blue)((4x+3))color(brown)((4x+3))`

`color(brown)(color(blue)(4x)(4x+3)color(blue)(+3)(4x+3))`

`16x^2+12x" "+12x+9" "=" "color(green)(16x^2+24x+9)`

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Substituting the expanded brackets and solving")`

`=>y=2x^2-15x-(color(green)(16x^2+24x+9))`

Multiply everything inside the brackets by -1 and group like terms

`=>y=2x^2-16x^2-15x-24x-9`

`=>y=-14x^2-29x-9`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From standard form we have:
`a=-14`
`b=-29`
`c=-9`

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

becomes`" "x=(+29+-sqrt((-29)^2-4(-14)(-9)))/(2(-14)`

`x=(29+-sqrt(841-504))/(-28)`

`x=(29+-sqrt(337))/(-28)" "larr 337" is a prime number"`

`color(blue)(x~~-1.69" and "x~~-0.38" to 2 decimal places")`