Akar akar persamaan kuadrat: 4x^2+px+25=0 adalah x1 dan x2. jika akar akar persamaan kuadrat x1^2+x2.... Question from @rahayufeb

October 2018 2 23 Report

Akar akar persamaan kuadrat: 4x^2+px+25=0 adalah x1 dan x2. jika akar akar persamaan kuadrat x1^2+x2^2=12,5, maka nilai p adalah

ImmanuelSinaga X1 + x2 = -b/a = -p/4
x1. x2 = c/a = 25/4

(x1 + x2)² - 2x1x2 = 12,5
(-p/4)² - 2(25/4) = 12,5
(-p/4)² = 12,5 + 25/2 = 25/2 + 25/2 = 25
(-p/4) = +√25 = +5

P1 = 5 x -4 = -20
P2 = -5 x -4 = 20

selamat belajar yh ^_^

0 votes Thanks 4

rahayufeb mkasih y

rahayufeb (x1+x2)^2-2x1x2=12,5 ....-2x1x2nya dari mana y

MicoArrafi Itu rumusnya

rahayufeb ooh mkasih y..dh ngingetin

MicoArrafi $$\begin{align} 4x^2+px+25=0\\\\x_1+x_2=-\frac{p}{4}\\x_1\times x_2=\frac{25}{4} \end{align}$

$$\begin{align} (x_1)^2+(x_2)^2&=(x_1+x_2)^2-2(x_1x_2)\\12,5&=\left(-\frac{p}{4}\right)^2-2\times\frac{25}{4}\\\frac{25}{2}&=\left(-\frac{p}{4}\right)^2-\frac{25}{2}\\\left(-\frac{p}{4}\right)^2&=\frac{25}{2}+\frac{25}{2}\\\left(-\frac{p}{4}\right)^2&=25\\-\frac{1}{4}p&=\pm5\\\\p_1=-5\times4=-20\\p_2=-5\times-4=20 \end{align}$

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