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Ditanya jika m+2 & n+2 akar2 persamaan kuadrat ax^2 + qx + r = 0, maka q+r ?
Jawab:
Note:
D = b2 - 4ac
m+n = = {(-b + √D) / 2a} + {(-b - √D) / 2a}
= (-b + √D - b - √D) / 2a
= -2b / 2a
= -b /a
m . n = = {(-b + √D) / 2a} {(-b - √D) / 2a}
= (b2 - D) / 4a2
= b2 - (b2 - 4ac) / 4a2
= (b2 - b2 + 4ac) / 4a2
= 4ac / 4a2
= c/a
Persamaan kuadrat dari akar2 m+2 & n+2:
ax^2 + ((m+2)+(n+2)) x + (m+2) (n+2) = 0
ax^2 + (m+n+4) x + (mn + 2m + 2n + 4) = 0
ax^2 + (m+n+4) x + (mn + 2 (m+n) + 4) = 0
ax^2 + qx + r = 0
jadi
q = m+n+4 = (-b/a)+4
r = mn + 2 (m+n) + 4 = (c/a) + 2 (-b/a) +4
q+r = ((-b/a) + 4) + ((c/a) - 2 (b/a) + 4)
= (c/a) - 3 (b/a) + 8