Solve the given quadratic equation by factorisation method, $(2 x - 3) = \sqrt{2 x^{2} - 2 x + 21}$

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Solution

The given equation $(2 x - 3) = \sqrt{2 x^{2} - 2 x + 21}$ can be simplified as follows:

$(2 x - 3) = \sqrt{2 x^{2} - 2 x + 21} \Rightarrow (2 x - 3)^{2} = (\sqrt{2 x^{2} - 2 x + 21})^{2} \Rightarrow (2 x)^{2} + 3^{2} - (2 \times 2 x \times 3) = 2 x^{2} - 2 x + 21 (∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b) \Rightarrow 4 x^{2} + 9 - 12 x = 2 x^{2} - 2 x + 21 \Rightarrow 4 x^{2} + 9 - 12 x - 2 x^{2} + 2 x - 21 = 0 \Rightarrow 2 x^{2} - 10 x - 12 = 0 \Rightarrow 2 (x^{2} - 5 x - 6) = 0 \Rightarrow x^{2} - 5 x - 6 = 0$

Now, the quadratic equation $x^{2} - 5 x - 6 = 0$ can be factorised as shown below:

$x^{2} - 5 x - 6 = 0 \Rightarrow x^{2} - 6 x + x - 6 = 0 \Rightarrow x (x - 6) + 1 (x - 6) = 0 \Rightarrow (x + 1) = 0, (x - 6) = 0 \Rightarrow x = - 1, x = 6$

Hence, $x = - 1$ or $x = 6$ .

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