Answer:
The solutions for the equation [tex]3x^{2} =7x-3[/tex] are: [tex]x_{1}=\frac{7+\sqrt{13}}{6} , x_{2}= \frac{7-\sqrt{13} }{6}[/tex], so the option d is correct.
Step-by-step explanation:
You have two options to find the solutions of the equation [tex]3x^{2} =7x-3[/tex]
The first is to solve it with the quadratic formula, for this, you need to follow these steps:
- Add 3 and substract 7x from both sides [tex]3x^{2} +3-7x=7x+3-3-7x[/tex]
- Simplify [tex]3x^{2} +3-7x=0[/tex]
- Use the quadratic equation formula. For a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] the solutions are [tex]x_{1,2} = \frac{-b\±\sqrt{b^{2}-4ac } }{2a}[/tex]
- In the equation given [tex]a=3, b=-7, c=3[/tex], so [tex]x_{1} = \frac{-(-7)+\sqrt{(-7)^{2}-4*3*3 } }{2*3}, x_{2} = \frac{-(-7)-\sqrt{(-7)^{2}-4*3*3 } }{2*3}[/tex]
- The solutions are [tex]x_{1}=\frac{7+\sqrt{13}}{6} , x_{2}= \frac{7-\sqrt{13} }{6}[/tex]
The second option is solve the equation by completing the square:
- Substract 7x from both sides [tex]3x^{2} -7x=7x-7x-3[/tex] and simplify [tex]3x^{2} -7x=-3[/tex]
- Divide both sides by 3 [tex]\frac{3x^{2}-7x }{3}= -\frac{3}{3}[/tex] and simplify [tex]x^{2} -\frac{7x}{3}=-1[/tex]
- Use the fact that [tex]x^{2} +2ax+a^{2}=(x+a)^{2}[/tex]
- We need to find [tex]a[/tex], for this, we use this relation [tex]2ax=-\frac{7x}{3}[/tex] and solve for [tex]a[/tex], we get [tex]a = -\frac{7}{6}[/tex]
- Add [tex]a^{2} = (\frac{-7}{6} )^{2} =\frac{13}{36}[/tex] to both sides in the equation of step 2 [tex]x^{2} -\frac{7x}{3}+(-\frac{7}{6} )^{2}=-1 +(-\frac{7}{6} )^{2}[/tex] and simplify [tex]x^{2} -\frac{7x}{3}+(-\frac{7}{6} )^{2}=\frac{13}{36}[/tex]
- Complete the square, use step 3, [tex]x^{2} -\frac{7x}{3}+(-\frac{7}{6} )^{2} = (x-\frac{7}{6} )^{2}=\frac{13}{36}[/tex]
- For [tex]f^{2}(x)[/tex] the solutions are [tex]f(x) =\sqrt{a},-\sqrt{a}[/tex], so [tex]x-\frac{7}{6} = \sqrt{\frac{13}{36} } =\frac{\sqrt{13}+7 }{6}\ and \ x-\frac{7}{6} = -\sqrt{\frac{13}{36} } =-\frac{\sqrt{13}+7 }{6}[/tex]
