Problem

Use the quadratic formula to solve for $x$
\[
6 x^{2}-9 x+1=0
\]

Answer

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Answer

Finally, rounding to three decimal places, the solutions to the equation \(6x^{2} - 9x + 1 = 0\) are \(x = \boxed{1.379}\) and \(x = \boxed{0.121}\).

Steps

Step 1 :Given the quadratic equation \(6x^{2} - 9x + 1 = 0\), we are asked to find the solutions for \(x\).

Step 2 :We can use the quadratic formula, which is given by \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation \(ax^{2} + bx + c = 0\).

Step 3 :In this case, \(a = 6\), \(b = -9\), and \(c = 1\). We can substitute these values into the quadratic formula to find the solutions for \(x\).

Step 4 :First, calculate the discriminant \(D\), which is \(b^{2} - 4ac\). Substituting the given values, we get \(D = 57\).

Step 5 :Next, substitute \(a\), \(b\), and \(D\) into the quadratic formula to find the solutions for \(x\). The solutions are \(x1 = 1.379152869605896\) and \(x2 = 0.12084713039410418\).

Step 6 :Finally, rounding to three decimal places, the solutions to the equation \(6x^{2} - 9x + 1 = 0\) are \(x = \boxed{1.379}\) and \(x = \boxed{0.121}\).

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