How do you find the roots, real and imaginary, of #y= -3x^2-4x+(2x- 1 )^2 # using the quadratic formula?
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"A 5.00 L sample of helium at STP expands to 15.0 L. What is the new pressure on the gas?"
# 4 +- [ (15)^(1/2) ] / 2#
If you expand the right side you should get
#x^2 - 8x +1#
Then just plug in to
# x = ( -b +- [b^2 - 4ac]^(1/2) ) / (2a) #
where a = 1 b = -8 c = 1
#x=4+sqrt15,##4-sqrt15#
Refer to the explanation for the process.
Given:
#y=-3x^2-4x+(2x-1)^2#
FOIL #(2x-1)^#
#y=-3x^2-4x+[4x^2-4x+1]#
Gather like terms.
#y=(-3x^2+4x^2)-(4x-4x)+1#
Combine like terms.
#y=x^2-8x+1# #larr# standard form: #y=ax^2+bx+c#,
where:
#a=1#, #b=-8#, and #c=1#
Substitute #0# for #y# and solve for #x# using the quadratic equation.
#0=x^2-8x+1#
Quadratic Formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Plug in the given values.
#x=(-(-8)+-sqrt((-8)^2-4*1*1))/(2*1)#
Simplify.
#x=(8+-sqrt(64-4))/2#
#x=(8+-sqrt60)/2#
Prime factorize #60#.
#x=(8+-sqrt(2xx2xx3xx15))/2#
Simplify.
#x=(8+-2sqrt15)/2#
Simplify.
#x=(color(red)cancel(color(black)(8^4))+-color(red)cancel(color(black)(2^1))sqrt15)/color(red)cancel(color(black)(2^1))#
#x=4+-sqrt15#
Roots
#x=4+sqrt15,##4-sqrt15#
