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Completing the square 4x^2 -11x+3=0
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To solve the quadratic equation 4x2−11x+3=04x^2 - 11x + 3 = 0 by completing the square, follow these steps:

Step 1: Divide by the Coefficient of x2x^2

First, divide the whole equation by 4 to make the coefficient of x2x^2 equal to 1.

x2−114x+34=0x^2 - \frac{11}{4}x + \frac{3}{4} = 0

Step 2: Move the Constant to the Right Side

x2−114x=−34x^2 - \frac{11}{4}x = -\frac{3}{4}

Step 3: Complete the Square

Take half of the coefficient of xx (which is $ -\frac{11}{4} $), square it, and add it to both sides.

(−118)2=12164\left(\frac{-11}{8}\right)^2 = \frac{121}{64}

Adding this to both sides:

x2−114x+12164=−34+12164x^2 - \frac{11}{4}x + \frac{121}{64} = -\frac{3}{4} + \frac{121}{64}

Convert −34-\frac{3}{4} to have a denominator of 64:

−34=−4864-\frac{3}{4} = -\frac{48}{64}

So,

−4864+12164=7364-\frac{48}{64} + \frac{121}{64} = \frac{73}{64}

Step 4: Write as a Square

(x−118)2=7364\left( x - \frac{11}{8} \right)^2 = \frac{73}{64}

Step 5: Solve for xx

Take the square root of both sides:

x−118=±738x - \frac{11}{8} = \pm \frac{\sqrt{73}}{8}

Step 6: Isolate xx

x=118±738x = \frac{11}{8} \pm \frac{\sqrt{73}}{8}

x=11±738x = \frac{11 \pm \sqrt{73}}{8}

So the solutions are:

x=11+738orx=11−738x = \frac{11 + \sqrt{73}}{8} \quad \text{or} \quad x = \frac{11 - \sqrt{73}}{8}