Question

factorize. x^2-1/4x-1

no comments please ............

Answer 1

Answer:

$(x - \frac{1 + \sqrt{65}}{8})(x - \frac{1 - \sqrt{65}}{8})$

Step-by-step explanation:

$Let\ \ x^2 - \frac{1}{4}x - 1 = 0 \\ \\ \\ a = 1\ \ ;\ \ b = -\frac{1}{4}\ \ ;\ \ c = -1 \\ \\ \\ b^2 - 4ac \\ \\ = (-\frac{1}{4})^2 - (4 \times 1 \times -1) \\ \\ = \frac{1}{16} - (-4) \\ \\ = \frac{1}{16} + 4\ =\ 4\frac{1}{16}\ =\ \frac{65}{16}$

$\\ \\ \\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \\ = \frac{\frac{1}{4} \pm \frac{\sqrt{65}}{4}}{2} \\ \\ = \frac{\frac{1 \pm \sqrt{65}}{4}}{2} \\ \\ = \frac{1 + \sqrt{65}}{8}\ \ \ \ \ OR\ \ \ \ \ \frac{1 - \sqrt{65}}{8}$

$\\ \\ \\ \therefore\ x^2 - \frac{1}{4}x - 1 = (x - \frac{1 + \sqrt{65}}{8})(x - \frac{1 - \sqrt{65}}{8})$

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