Solve the following equation by factorization:
x2−(1+2)x+2x^2 - (1 + \sqrt{2})x + \sqrt{2}x2−(1+2)x+2 = 0
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Given,
⇒x2−(1+2)x+2=0⇒x2−1x−2x+2=0⇒x(x−1)−2(x−1)=0⇒(x−2)(x−1)=0⇒(x−2)=0 or (x−1)=0 [Using Zero-product rule] ⇒x=2 or x=1.\Rightarrow x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0 \\[1em] \Rightarrow x^2 - 1x - \sqrt{2}x + \sqrt{2} = 0 \\[1em] \Rightarrow x(x - 1) - \sqrt{2}(x - 1) = 0 \\[1em] \Rightarrow (x - \sqrt{2})(x - 1) = 0 \\[1em] \Rightarrow (x - \sqrt{2}) = 0 \text{ or } (x - 1) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow x = \sqrt{2} \text{ or } x = 1.⇒x2−(1+2)x+2=0⇒x2−1x−2x+2=0⇒x(x−1)−2(x−1)=0⇒(x−2)(x−1)=0⇒(x−2)=0 or (x−1)=0 [Using Zero-product rule] ⇒x=2 or x=1.
Hence, x={1,2}x = {1, \sqrt{2}}x={1,2}.
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