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

User Momer
by
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2 Answers

9 votes

Answer:

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

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)

User Chen Kinnrot
by
7.6k points
0 votes

Answer:


\displaystyle (7 + √(85))/(6) and
\displaystyle (7 - √(85))/(6).

Explanation:

(Replace
b with
x to avoid confusion with symbols in the quadratic equation.)

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


3\, x^(2) - 7\, x + 3 = 6.


3\, x^(2) - 7\, x + 3 - 6 = 0.


3\, x^(2) + (- 7)\, x + (-3) = 0.

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


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


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

User Ekfuhrmann
by
7.7k points

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