If the zeros of #3x^2+5x-2# are #alpha# and #beta#, then what quadratic has zeros #2alpha# and #2beta# ?

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Answer 1 · 2017-12-24T16:38:42

#3x^2+10x-8=0#

We can factorise #3x^2+5x-2=0# to get #(3x-1)(x+2)=0#

For this to equal 0, either #(3x-1)# or #(x+2)# must equal 0.

#3x-1=0#
#3x=1#
#x=1/2#

#x-2=0#
#x=2#

We will call #1/3# #alpha#, and #-2# #beta#.

#2alpha=2/3#, #2beta=-4#

So, #x+4=0#, or #x=2/3 => 3x=2 => 3x-2=0#

#(3x-2)(x+4)=0#

By expanding, we get:
#3x^2-2x+12x-8=0#

#3x^2+10x-8=0#

Answer 2 · 2017-12-24T19:06:50

#3x^2+10x-8#

Substituting #x = t/2# then we get:

#3x^2+5x-2 = 3(t/2)^2+5(t/2)-2#

#color(white)(3x^2+5x-2) = 3/4t^2+5/2t-2#

Multiplying through by #4#, we could specify the polynomial as:

#3t^2+10t-8#

or using #x# instead of #t#

#3x^2+10x-8#