Question

tentukan akar² persamaan kuadrat dari:
(x + 1)² + (x - 1)² = (2x + 1)² - (2x - 1)²​

Answer 1

Jawab:

2 + √3 atau 2 - √3

Penjelasan dengan langkah-langkah:

Jabarkan

[tex]\begin{aligned}(x+1)^2+(x-1)^2&=(2x+1)^2-(2x-1)^2\\x^2+2x+1+x^2-2x+1&=4x^2+4x+1-(4x^2-4x+1)\\2x^2+2&=8x\\2x^2-8x+2&=0\end{aligned}[/tex]

Selesaikan dengan rumus kuadratik

[tex]\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-8)\pm\sqrt{(-8)^2-4(2)(2)}}{2(2)}\\&=\frac{8\pm\sqrt{64-16}}{4}\\&=\frac{8\pm 4\sqrt{3}}{4}\\&=2\pm\sqrt{3}\end{aligned}[/tex]

Answer 2

Verified answer

Penjelasan

[tex]\tt (x+1)^2+(x-1)^2=(2x+1)^2-(2x-1)^2\\\\x^2+2x+1+x^2-2x+1=(2x+1-2x+1)(2x+2x)\\\\\frac{2x^2+2}{2} =\frac{8x}{2} \\\\x^2+1=4x\\\\x^2-4x+1=0\\\\a=1\\\\b=-4\\\\c=1\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\x_{1,2}=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}\\ \\x_{1,2}=\frac{4\pm 2\sqrt{3}}{2}\\ \\x_1=\frac{4-2\sqrt{3}}{2}\\ \\x_1=\frac{2(2-\sqrt{3})}{2}\\ \\x_1=2-\sqrt{3}\\ \\\\x_2=\frac{4+2\sqrt{3}}{2}\\ \\x_2=\frac{2(2+\sqrt{3})}{2}\\ \\x_2=2+\sqrt{3}[/tex]

Hasil dari akar² persamaan kuadrat dari (x + 1)² + (x - 1)² = (2x + 1)² - (2x - 1)²​ adalah x₁ = 2-√3, x₂ = 2 + √3