jjm12
jjm12 jjm12
  • 05-12-2018
  • Mathematics
contestada

Q.16 please
With steps

Q16 please With steps class=

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LammettHash
LammettHash LammettHash
  • 05-12-2018

The equations

[tex]2x^2+9x-5=0[/tex]

and

[tex]2\left(t-\dfrac12\right)^2+9\left(t-\dfrac12\right)-5=0[/tex]

are really the same, we're just setting [tex]x=t-\dfrac12[/tex], or [tex]t=x+\dfrac12[/tex].

[tex]2x^2+9x-5=(2x-1)(x+5)=0\implies x=\dfrac12,x=-5[/tex]

So we get

[tex]t=\dfrac12+\dfrac12=1,t=-5+\dfrac12=-\dfrac92[/tex]

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