5(x²+2)=2x(3x+1)
Tlng....
Mohon di jawab dgn lengkap... persamaan kuadrat
Jawab:
[1.] 2x(3x+5) = 4
[tex]\sf Hp=\left[\frac{1}{3}~~,~~-2\right][/tex]
[2.] 5(x²+2)=2x(3x+1)
Hp = {√11 - 1 , -√11 - 1}
Penjelasan:
[1.] 2x(3x+5) = 4
2x(3x) + 2x(5) = 4
6x² + 10x = 4
6x² + 10x - 4 = 0
[tex]\frac{6x^2}{6}+\frac{10x}{6}-\frac{4}{6}=0\\\\x^2+\frac{10x\div2}{6\div2}-\frac{4\div2}{6\div2}=0\\\\x^2+\frac{5x}{3}-\frac{2}{3}=0\\\\\because x^2+bx+c=0\\\rightarrow x^2+bx+(\frac{1}{2}\cdot b)^2=(\frac{1}{2}\cdot b)^2-c\\\\\therefore x^2+\frac{5x}{3}-\frac{2}{3}=0\\\\\rightarrow x^2+\frac{5x}{3}+(\frac{1}{2}\cdot \frac{5}{3})^2=(\frac{1}{2}\cdot \frac{5}{3})^2+\frac{2}{3}\\\\x^2+\frac{5x\cdot2}{3\cdot2}+(\frac{5}{6})^2=(\frac{5}{6})^2+\frac{2}{3}[/tex]
[tex]x^2+\frac{10x}{6}+(\frac{5}{6})^2=\frac{5^2}{6^2}+\frac{2\cdot12}{3\cdot12}\\\\x^2+2x(\frac{5}{6})+(\frac{5}{6})^2=\frac{25}{36}+\frac{24}{36}\\\\(~~a=x,~~b=\frac{5}{6}~~)\\\\\because a^2+2ab+b^2=(a+b)^2\therefore\\\\(x+\frac{5}{6})^2=\frac{25}{36}+\frac{24}{36}\\\\(x+\frac{5}{6})^2=\frac{49}{36}\\\\x+\frac{5}{6}=\pm\sqrt{\frac{49}{36}}\\\\x+\frac{5}{6}=\pm\frac{\sqrt{49}}{\sqrt{36}}\\\\x+\frac{5}{6}=\pm\frac{7}{6}\\\\x=\pm\frac{7}{6}-\frac{5}{6}[/tex]
[tex]\sf Hp=\left[\frac{7}{6}-\frac{5}{6}~~,~~-\frac{7}{6}-\frac{5}{6}\right]\\\\Hp=\left[\frac{2}{6}~~,~~-\frac{12}{6}\right]\\\\Hp=\left[\frac{2\div2}{6\div2}~~,~~-\frac{12}{6}\right]\\\\\large\boxed{Hp=\left[\frac{1}{3}~~,~~-2\right]}[/tex]
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[2.] 5(x²+2)=2x(3x+1)
5x² + 5(2) = 2x(3x) + 2x(1)
5x² + 10 = 6x² + 2x
6x² - 5x² + 2x - 10 = 0
x² + 2x - 10 = 0
x² + 2x = 10
x² + 2x + (½·2)² = 10 + (½·2)²
x² + 2x + 1² = 10 + 1²
x² + 2x(1) + 1² = 10 + 1
(x+1)² = 11
x+1 = ±√11
x = ±√11 - 1
Hp = {√11 - 1 , -√11 - 1}
(xcvi)
[answer.2.content]
