Persiapan Kuis Matematika [Persamaan Kuadrat]

hanya (1)

hanya (2)

hanya (3)

(1) dan (2)

(1), (2) dan (3)

LINEAR DUA VARIABEL

KUADRAT

LINEAR SATU VARIABEL

LINEAR TIGA VARIABEL

POLINOMIAL

a = 36 ; b = 12 ; c = 1

a = 36 ; b = -12 ; c = 1

a = -1 ; b = -12 ; c = 36

a = 1 ; b = 12 ; c = -36

a = 1 ; b = -12 ; c = 36

a = 3 ; b = -11 ; c = 21

a = 1 ; b = 0 ; c = 7

a = 3 ; b = 11 ; c = 21

a = 2 ; b = 11 ; c = 14

a = 2 ; b = -11 ; c = 14

x₁ = 6 atau x₂ = 16

x₁ = -2 atau x₂ = 8

x₁ = 6 atau x₂ = -16

x₁ = -2 atau x₂ = -8

x₁ = 2 atau x₂ = -8

 $p_{1\ }=\ \frac{\text{5}}{\text{2}}\ atau\ p_{2\ }=\ 2$  

 $p_{1\ }=\ -\frac{1}{2}\ atau\ p_{2\ }=\ -10$  

 $p_{1\ }=-\ \frac{5}{\text{2}}\ atau\ p_{2\ }=\ 2$  

 $p_{1\ }=\ \frac{\text{1}}{\text{2}}\ atau\ p_{2\ }=\ -10$  

 $p_{1\ }=\ -\frac{\text{5}}{\text{2}}\ atau\ p_{2\ }=\ -2$  

 $\left[n-3\right]^2\ =\ -\frac{3}{2}$  

 $\left[n+3\right]^2\ =\ \frac{3}{2}$  

 $\left[n+3\right]^2\ =\ \frac{15}{2}$  

 $\left[n-3\right]^2\ =\ -\frac{15}{2}$  

 $\left[n-3\right]^2\ =\frac{15}{2}$  

 $\left[x-3\right]^2\ =\ 1$  

 $\left[x+3\right]^2\ =\ 1$  

 $\left[x\pm3\right]^2\ =\ 1$  

 $\left[x+3\right]^2\ =\ -1$  

 $\left[x-3\right]^2\ =\ -1$  

 $m_{\left(1,2\right)}\ =\ \frac{3\pm\sqrt{-17\ }}{4}$  

 $m_{\left(1,2\right)\ }=\ \frac{-3\pm\sqrt{17}}{4}$  

 $m_{\left(1,2\right)}=\frac{3\pm\sqrt{5}}{4}$  

 $m_{\left(1,2\right)}\ =\ \frac{-3\pm\sqrt{-17}}{4}$  

 $m_{\left(1,2\right)}=\ \frac{3\pm\sqrt{17}}{4}$  

 $z_{\left(1,2\right)}\ =\ 4\ \pm\sqrt{10}$  

 $z_{\left(1,2\right)}\ =\ -4\ \pm\ \sqrt{10}$  

 $z_{\left(1,2\right)}\ =\ 4\ \pm\sqrt{40}$  

 $z_{\left(1,2\right)}\ =-\ 4\ \pm\ \sqrt{40}$  

 $z_{\left(1,2\right)}\ =\ -4\ \pm\ 2\sqrt{-10}$