Jika a dan b adalah akar-akar persamaan kuadrat 2x^(2)−6x+3=0, maka nilai dari ab^(2)+a^(2)b adalah... A. 4 B. 5 C. 4,5 D. 5.5

<p>Halo Muhammad. Kakak bantu jawab ya.</p><p>Jawab: C</p><p>Pembahasan:</p><p>Ingat!</p><p>Jika diketahui persamaan kuadrat ax²+bx+c=0, rumus kuadratik untuk menentukan akar-akar persamaan kuadrat adalah&nbsp;</p><p>x_1,2=(-b±√D)/2a dengan D=b²-4ac.</p><p>Dari persamaan kuadrat 2x²−6x+3=0 diperoleh bahwa a=2, b=-6, c=3. Sehingga,</p><p>D = b²-4ac</p><p>&nbsp; &nbsp; = (-6)²-4(2)(3)</p><p>&nbsp; &nbsp; = 36-24</p><p>&nbsp; &nbsp; = 12</p><p>Oleh karena itu, akar-akar persamaan kuadratnya adalah</p><p>a,b = (-(-6)±√12)/(2(2))</p><p>&nbsp; &nbsp; &nbsp; &nbsp;= (6±2√3)/4</p><p>a = (6+2√3)/4 &nbsp; &nbsp; &nbsp; &nbsp;atau &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;b = (6-2√3)/4</p><p>&nbsp; &nbsp; = (3+√3)/2 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = (3-√3)/2</p><p>Perhatikan bahwa</p><p>b² = ((3-√3)/2)²</p><p>&nbsp; &nbsp; &nbsp; = (12-6√3)/4</p><p>&nbsp; &nbsp; &nbsp; = (6-3√3)/2</p><p>a² = ((3+√3)/2)²</p><p>&nbsp; &nbsp; &nbsp; = (12+6√3)/4</p><p>&nbsp; &nbsp; &nbsp; = (6+3√3)/2</p><p>Sehingga,</p><p>ab²= ((3+√3)/2 )×( (6-3√3)/2)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; = (9-3√3)/4</p><p>a²b = ((6+3√3)/2)×((3-√3)/2)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; = &nbsp;(9+3√3)/4</p><p>Oleh karena itu,&nbsp;</p><p>ab² + a²b = (9-3√3)/4 + (9+3√3)/4</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 18/4</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 4,5</p><p>&nbsp;</p><p><strong><u>Jadi, jawaban yang benar adalah C.</u></strong></p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p>