SOLUTION: Let r and s be the roots of y^2 - 19y + 7. Find (r-2)(s-2).

Algebra.Com
Question 1076467: Let r and s be the roots of y^2 - 19y + 7. Find (r-2)(s-2).

Found 2 solutions by rothauserc, MathTherapy:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
y^2 -19y + 7 = 0
:
complete the square
:
y^2 -19y +(361/4) = -(28/4) +(361/4)
:
(y -19/2)^2 = 333/4
:
y = (19/2) +(square root(333)/2) = (37.2463/2) = 18.6231
:
y = (19/2) -(square root(333)/2) = (.7537/2) = .3768
:
let r = 18.6231 and s = .3768
:
**********************************************
(r-2)(s-2) = (18.6231-2)(.3768-2) = −26.9826
**********************************************
:

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Let r and s be the roots of y^2 - 19y + 7. Find (r-2)(s-2).
(r – 2)(s – 2)_____rs – 2r – 2s + 4______rs – 2(r + s) + 4
Product, or
Sum, or
rs – 2(r + s) + 4
7 – 2(19) + 4 -------- Substituting 7 for rs, and 19 for r + s

It's as simple as that! No complex calculations are necessary!
RELATED QUESTIONS

let the roots of the equation x^3 -2x^2 -3x-7=0 be r, s, and t. find the equation whose... (answered by math_tutor2020)
R is inversel;y proportinal to the cube of S. Let k be the constant of proportionally. If (answered by stanbon)
Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 +... (answered by ikleyn)
If r and s are the roots of x^2 -8x +6 =0, find r^2 + 3rs +s^2. Thank... (answered by solver91311,ikleyn,MathTherapy)
Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 +... (answered by ikleyn,Edwin McCravy,mccravyedwin)
Let r_1, r_2, r_3, r_4, and r_5 be the complex roots of x^5 - 4x^2 + 7x - 1 = 0. Compute (answered by CPhill,ikleyn)
Given that r and s are roots of the quadratic equation 3(x^2)+1=7x, find (r^3)s+r(s^3),... (answered by drk)
Let r(x)=x^2-3 and s(x)=x^3-6. Find... (answered by jim_thompson5910)
Determine {{{(r + s)(s + t)(t + r)}}}, if r, s, and t are the three real roots of the... (answered by ikleyn)