Nilai Diskriminan

\(x^2 \:-\: 12x + 35 = 0\)

\(a = 1\)

\(b = -12\)

\(c = 35\)

Nilai diskriminan \(\text{D} = b^2 \:-\: 4ac\)

\(\text{D} = (-12)^2 \:-\: 4(1)(35)\)

\(\text{D} = 144 \:-\:140 = 4 > 0\)

Karena \(\text{D} > 0\), maka persamaan kuadrat \(x^2 \:-\:12x + 35 = 0\) memiliki dua akar real yang berbeda

Jika persamaan kuadrat \(x^2 \:-\:12x + 35 = 0\) kita faktorkan akan menjadi\((x\:-\:5)(x\:-\:7)=0\)

\(x \:-\: 5 = 0 \rightarrow x_1 = 5\)

\(x\:-\:7 = 0 \rightarrow x_2 = 7\)

Akar-akar realnya adalah 5 dan 7

\(x^2\:-\: 16x + 64 = 0\)

\(a = 1\)

\(b = -16\)

\(c = 64\)

Nilai diskriminan \(\text{D} = b^2 \:-\:4ac\)

\(\text{D} = (-16)^2 \:-\: 4(1)(64)\)

\(\text{D} = 256 \:-\: 256 = 0\)

Karena \(\text{D} = 0\), maka persamaan kuadrat \(x^2 \:-\: 16x + 64 = 0\) memiliki akar yang kembar

Jika persamaan kuadrat \(x^2 \:-\:16x + 64 = 0\) kita faktorkan akan menjadi\((x\:-\:8)(x\:-\:8)=0\)

\(x \:-\: 8 = 0 \rightarrow x_1 = x_2 = 8\)

Akar kembarnya adalah 8

\(x^2 +4x + 10 = 0\)

\(a = 1\)

\(b = 4\)

\(c = 10\)

Nilai diskriminan \(\text{D} = b^2\:-\: 4ac\)

\(\text{D} = 4^2 \:-\: 4(1)(10)\)

\(\text{D} = 16 \:-\: 40 = -24 < 0\)

Karena \(\text{D} < 0\), maka persamaan kuadrat \(x^2 +4x + 10 = 0\) tidak memiliki akar-akar real (memiliki akar-akar imajiner)

Dengan metode kuadrat sempurna akan kita cari akar-akar imajinernya

\(x^2 +4x + 10 = 0\)

\(x^2 + 4x = -10\)

\((x + 2)^2 \:-\: 2^2 = -10\)

\((x+ 2)^2 = -10 + 4\)

\((x + 2)^2 = -6\)

\(x + 2 = \pm \sqrt{-6}\:\:\:\:\:\color{blue} \sqrt{-1} = i\)

\(x = -2 \pm \sqrt{6} i\)

Akar-akar imajinernya adalah \(-2 + \sqrt{6} i\) dan \(-2 \:-\: \sqrt{6} i\)

\(x^2 + (2m\:-\:1)x + m^2 \:-\: 3m + 5 = 0\)

\(a = 1\)

\(b = 2m\:-\:1\)

\(c = m^2\:-\: 3m + 5\)

Agar persamaan kuadrat memiliki akar-akar real maka \(\text{D} \geq 0\)

\(b^2 \:-\: 4ac \geq 0\)

\((2m\:-\:1)^2 \:-\:4(1)(m^2 \:-\:3m + 5) \geq 0\)

\(4m^2 \:-\:4m + 1 – 4m^2 + 12 m \:-\: 20 \geq 0\)

\(  8m \:-\: 19 \geq 0\)

\(m \geq \dfrac{19}{8}\)

\((k+4)x^2 \:-\: k\sqrt{2}x + k \:-\: 3 = 0\)

\(a = k+4\)

\(b = -k\sqrt{2}\)

\(c = k\:-\:3\)

Agar persamaan kuadrat memiliki dua akar real yang berbeda maka \(\text{D} > 0\)

\(b^2\:-\:4ac > 0\)

\((-k\sqrt{2})^2 \:-\:4(k+4)(k\:-\:3) > 0\)

\(2k^2\:-\:4(k^2 \:-\:3k + 4k – 12) > 0\)

\(2k^2\:-\:4(k^2 + k – 12) > 0\)

\(2k^2\:-\: 4k^2\:-\:4k +48 > 0\)

\(-2k^2 \:-\:4k +48 > 0\)

\(-2(k^2 +2k \:-\:24) > 0\)

\(-2(k+6)(k\:-\:4) > 0\)

\(k+6 = 0 \rightarrow k = -6\)

\(k\:-\:4 = 0 \rightarrow k = 4\)

Jadi batasan nilai \(k\) nya adalah \(-6 < k < 4\)

\(x^2 +(m + 1)x + 2 \:-\: m = 0\)

\(a = 1\)

\(b = m + 1\)

\(c = 2\:-\:m\)

Agar persamaan kuadrat tidak memiliki akar-akar real maka \(\text{D} < 0\)

\(b^2 \:-\: 4ac < 0\)

\((m + 1)^2\:-\:4(1)(2- m) < 0\)

\(m^2 + 2m + 1\:-\: 8 + 4m < 0\)

\(m^2 + 6m\:-\:7 < 0\)

\((m + 7)(m \:-\: 1) < 0\)

\(m + 7 = 0 \rightarrow m = -7\)

\(m – 1 = 0 \rightarrow m = 1\)

Jadi batasan nilai \(m\) nya adalah \(-7 < m < 1\)

\((m+1)x^2\:-\: 2mx + m\:-\: 2 = 0\)

\(a = m+1\)

\(b = -2m\)

\(c = m-2\)

Agar persamaan kuadrat memiliki akar kembar maka \(\text{D} = 0\)

\((-2m)^2 \:-\:4(m+1)(m\:-\:2) = 0\)

\(4m^2 \:-\:4(m^2\:-\:2m + m – 2) = 0\)

\(4m^2 \:-\: 4(m^2\:-\: m \:-\: 2) = 0\)

\(4m^2 \:-\: 4m^2 +4m +8 = 0\)

\(4m = -8\)

\(m = -\dfrac{8}{4} = -2\)