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Solve the equation for all real solutions in simplest form.
3b^{2}-7b+3= 6


Sagot :

Answer:

b=(7-sqrd85/6), b=(7+sqrd85/6)

Step-by-step explanation:

3b^2-7b+3=6

3b^2-7b+3-6=0

3b^2-7b-3=0

a=3, b=-7, c=-3

b=7+-sqrd49+36/6

b=(7-sqrd85/6), b=(7+sqrd85/6)

Answer:

[tex]\displaystyle \frac{7 + \sqrt{85}}{6}[/tex] and [tex]\displaystyle \frac{7 - \sqrt{85}}{6}[/tex].

Step-by-step explanation:

(Replace [tex]b[/tex] with [tex]x[/tex] to avoid confusion with symbols in the quadratic equation.)

Notice that the equation [tex]3\, x^{2} - 7\, x + 3 = 6[/tex] is quadratic with respect to the unknown [tex]x[/tex]. Rewrite the equation in standard form [tex]a\, x^{2} + b\, x + c = 0[/tex] before applying the quadratic equation:

[tex]3\, x^{2} - 7\, x + 3 = 6[/tex].

[tex]3\, x^{2} - 7\, x + 3 - 6 = 0[/tex].

[tex]3\, x^{2} + (- 7)\, x + (-3) = 0[/tex].

Thus, for the quadratic equation, [tex]a = 3[/tex], [tex]b = (-7)[/tex], and [tex]c = (-3)[/tex]. Apply the quadratic equation to find the solutions:

[tex]\begin{aligned}x &= \frac{-\, b + \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(-7) + \sqrt{(-7)^{2} - 4 \times 3 \times (-3)}}{2 \times 3} \\ &= \frac{7 + \sqrt{85}}{6}\end{aligned}[/tex].

[tex]\begin{aligned}x &= \frac{-\, b - \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(-7) - \sqrt{(-7)^{2} - 4 \times 3 \times (-3)}}{2 \times 3} \\ &= \frac{7 - \sqrt{85}}{6}\end{aligned}[/tex].