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Solve for [tex]$x$[/tex].

Hint: Put the quadratic equation in standard form so that only a zero is on the right side.

[tex]3x^2 = 5x - 1[/tex]

A. [tex]x = \frac{-5 \pm \sqrt{13}}{6}[/tex]
B. [tex]x = \frac{5 \pm \sqrt{13}}{3}[/tex]
C. [tex]x = \frac{5 \pm \sqrt{13}}{6}[/tex]
D. [tex]x = \frac{5 \pm \sqrt{37}}{6}[/tex]


Sagot :

To solve the given quadratic equation:

[tex]\[ 3x^2 = 5x - 1 \][/tex]

we need to start by rewriting it in the standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex]. This involves bringing all terms to one side of the equation:

[tex]\[ 3x^2 - 5x + 1 = 0 \][/tex]

This standard form allows us to identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:

[tex]\[ a = 3, \][/tex]
[tex]\[ b = -5, \][/tex]
[tex]\[ c = 1. \][/tex]

The quadratic formula to find the roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \][/tex]

First, we calculate the discriminant ([tex]\( \Delta \)[/tex]):

[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 3 \cdot 1 = 25 - 12 = 13. \][/tex]

Next, we substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( \Delta \)[/tex] back into the quadratic formula:

[tex]\[ x = \frac{-(-5) \pm \sqrt{13}}{2 \cdot 3} = \frac{5 \pm \sqrt{13}}{6}. \][/tex]

This results in the two potential solutions for [tex]\( x \)[/tex]:

[tex]\[ x_1 = \frac{5 + \sqrt{13}}{6}, \][/tex]
[tex]\[ x_2 = \frac{5 - \sqrt{13}}{6}. \][/tex]

Thus, the correct form of the solutions corresponds to option C.

Therefore, the correct choice is:

C) [tex]\( x = \frac{5 \pm \sqrt{13}}{6} \)[/tex]