87. Jika selisih akar – akar x2 – nx + 24 = 0 persamaan sama dengan 5, maka jumlah akar – akar persamaan adalah

[tex]\displaystyle x_1-x_2=5\\\pm\frac{\sqrt{b^2-4ac}}{a}=5\\\pm\frac{\sqrt{(-n)^2-4(1)(24)}}{1}=5\\\pm\sqrt{n^2-96}=5\\n^2-96=25\\n^2=121\\n=\pm11\\\\x_1+x_2=-\frac{b}{a}\\x_1+x_2=-\frac{-n}{a}\\x_1+x_2=\frac{n}{a}\\x_1+x_2=\frac{\pm11}{1}\\\boxed{\boxed{x_1+x_2=\pm11}}[/tex]

Akar akar  persamaan kuadrat
x² - nx + 24 = 0
a =1 , b = - n, c = 24
x1+ x2 = -b/a = n
x1 x2  = c/a = 24

x1 - x2 = 5
(x1 - x2)²  = (5)²
(x1 + x2)² - 4(x1x2) = (5)²
(n)² - 4(24) = 25
n² = 25 + 96= 121
n= √121
n =11 atau n = - 11