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

Question

1585 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-30T20:50:21

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

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

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

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

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

Tony B

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