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# algebra

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Let r and s be the roots of y^2 - 19y + 9. Find (r-2)(s-2).

Apr 4, 2021

#1
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Here is a start:

(r-2)(s-2).= rs -2r-2s+4

=  rs -2 (r+s) +4    Are you familiar with Vieta's formulas for quadratics???

Vieta's for quadratics   :   r + s = -b/a   =19

and   rs = c/a     = 9

rs -2 (r+s) +4 =

9 -2(19)+4 = -25

Apr 4, 2021
edited by ElectricPavlov  Apr 4, 2021
edited by ElectricPavlov  Apr 4, 2021
#2
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For this, we don't ACTUALLY have to solve for r and s.

Let us first expand (r-2)(s-2):

$r\cdot{s}+r\cdot{-2}+{-2}\cdot{s}+{-2}\cdot{-2}=rs-2r-2s+4$

if r and s are the roots of y^2 - 19y + 9, then:

$y^2 - 19y + 9 = (y-r)(y-s) = y^2 - (r + s)y + rs$

so r+s=19 and rs=9

going back to the original equation, we have

$rs-2r-2s+4=9-2(r+s)+4=9-38+4=\boxed{-25}$

Apr 4, 2021
edited by SparklingWater2  Apr 4, 2021