Quadratic Equations – Part 7

Type 7:  2x^{2}-(4+\sqrt{13})x+2\sqrt{13}

Basic method : 2x^{2}-4x-\sqrt{13}x+2\sqrt{13}

2x(x-2)-\sqrt{13}(x-2)

(2x-\sqrt{13})(x-2)

2x-\sqrt{13}=0            x-2=0

x=\frac{\sqrt{13}}{2}                          x=2

Shortcut method:

Step 1: Consider the middle term  (4+\sqrt{13})

Step 2: Split the term into 4 and \sqrt{13}

Step 3: Divide  4 and \sqrt{13}  by the first term 2.

\frac{4}{2},\frac{\sqrt{13}}{2}=2,\frac{\sqrt{13}}{2}

The roots of the given equation are  2,\frac{\sqrt{13}}{2}

10y^{2}-(18+5\sqrt{13})y+9\sqrt{13}

Step 1: Consider the middle term  (18+5\sqrt{13})

Step 2: Split the term into 18 and 5\sqrt{13}

Step 3: Divide  18 and 5\sqrt{13}  by the first term 10

\frac{18}{10},\frac{5\sqrt{13}}{10}=1.8,\frac{\sqrt{13}}{2}

The roots of the given equation are  1.8,\frac{\sqrt{13}}{2}

The final answer is x\geq y

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