X²-2X-5 find the value of X by Factorisation method
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Hey friend, Harish here.
Here is your answer.
![x^2 - 2x - 5 = 0 \\ \\ \to x^2 -\Bigl[ (x - \sqrt{6}x) +(x + \sqrt{6}x) \Bigr ]+\bigl(1- \sqrt{6}\bigr)\bigl(1+\sqrt{6}\bigr) = 0 \\ \\ \to x\bigl(x - 1 +\sqrt{6}\bigr) - (1+ \sqrt{6} )(x-1+ \sqrt{6}) = 0\\ \\ \mathrm{Now \ taking\ (x-1+ \sqrt{6})\ as\ common ;}\\ \\ \to (x-1+ \sqrt{6})(x-1- \sqrt{6}) = 0\\ \\ \boxed{\mathrm{Therefore: x = 1+ \sqrt{6} , x = 1- \sqrt{6}}}](https://tex.z-dn.net/?f=x%5E2+-+2x+-+5+%3D+0+%5C%5C+%5C%5C+%5Cto+x%5E2+-%5CBigl%5B+%28x+-++%5Csqrt%7B6%7Dx%29+%2B%28x+%2B+%5Csqrt%7B6%7Dx%29+%5CBigr+%5D%2B%5Cbigl%281-+%5Csqrt%7B6%7D%5Cbigr%29%5Cbigl%281%2B%5Csqrt%7B6%7D%5Cbigr%29++%3D+0++%5C%5C+%5C%5C+%5Cto+x%5Cbigl%28x+-+1+%2B%5Csqrt%7B6%7D%5Cbigr%29+-+%281%2B+%5Csqrt%7B6%7D+%29%28x-1%2B+%5Csqrt%7B6%7D%29+%3D+0%5C%5C+%5C%5C+%5Cmathrm%7BNow+%5C+taking%5C+%28x-1%2B+%5Csqrt%7B6%7D%29%5C+as%5C+common+%3B%7D%5C%5C+%5C%5C++%5Cto+%28x-1%2B+%5Csqrt%7B6%7D%29%28x-1-+%5Csqrt%7B6%7D%29++%3D+0%5C%5C+%5C%5C+%5Cboxed%7B%5Cmathrm%7BTherefore%3A+x+%3D+1%2B+%5Csqrt%7B6%7D+%2C+x+%3D+1-+%5Csqrt%7B6%7D%7D%7D+ "x^2 - 2x - 5 = 0 \\ \\ \to x^2 -\Bigl[ (x - \sqrt{6}x) +(x + \sqrt{6}x) \Bigr ]+\bigl(1- \sqrt{6}\bigr)\bigl(1+\sqrt{6}\bigr) = 0 \\ \\ \to x\bigl(x - 1 +\sqrt{6}\bigr) - (1+ \sqrt{6} )(x-1+ \sqrt{6}) = 0\\ \\ \mathrm{Now \ taking\ (x-1+ \sqrt{6})\ as\ common ;}\\ \\ \to (x-1+ \sqrt{6})(x-1- \sqrt{6}) = 0\\ \\ \boxed{\mathrm{Therefore: x = 1+ \sqrt{6} , x = 1- \sqrt{6}}}")
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Hope my answer is helpful to you.
Here is your answer.
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Hope my answer is helpful to you.
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