Question

Tentukan Deskriminan, jenis akar dan penyelesaian dari persamaan kuadrat berikut!

(1) x²- 6x + 9 = 0
(2) x² - 2x + 13 = 0
(3) 2x² - 2x - 24 = 0​

Answer 1

Penjelasan dengan langkah-langkah:

DISKRIMINAN

= b²-4ac

1.

x² - 6x + 9 = 0

ax²+bx+c = 0

Diskriminan

D = b²-4ac

D = (-6)²-4(1)(9)

D = 36-36

D = 0

Jenis akar

Akar real kembar

Penyelesaian

x² - 6x + 9 = 0

(x-3)(x-3) = 0

x1 = 3 dan x2 = 3

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2.

x²-2x+13 = 0

ax²+bx+c = 0

Diskriminan

D = b²-4ac

D = (-2)²-4(1)(13)

D = 4-52

D = -48

Karena D < 0 maka

Jenis akar

Akar akar tidak real

Penyelesaian

x²-2x+13 = 0

[tex] = \frac{ - b\pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ = \frac{ - ( - 2)\pm \sqrt{( - 2 {)}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{2\pm \sqrt{-48} }{2} \\ \\ x1 = \frac{2 + \sqrt{-48} }{2} \\ = \sqrt{-48} \\ \\ x2 = \frac{2 - \sqrt{48} }{2} \\ = \sqrt{48} [/tex]

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3.

2x² - 2x - 24 = 0

ax²+bx+c = 0

Diskriminan

D = b²-4ac

D = (-2)²-4(2)(-24)

D = 4+192

D = 196

Karena D > 0

Jenis akar

Akar real tapi berbeda

Penyelesaian

2x²-2x+24 = 0

[tex] = \frac{ - b\pm \sqrt{D} }{2a} \\ = \frac{2\pm \sqrt{196} }{2(2)} \\ = \frac{2\pm14}{4} \\ \\ x1 = \frac{2 + 14}{4} \\ = 4 \\ \\ x2 = \frac{2 - 14}{4} \\ = - 3[/tex]