Question:
Show that the quadratic equation $x^{2}-8 x+18=0$ has no real solution.
Solution:
Given: $x^{2}-8 x+18=0$
$x^{2}-8 x+18=0$
Adding and subtracting $\left(\frac{1}{2} \times 8\right)^{2}$, we get
$\Rightarrow x^{2}-8 x+18+4^{2}-4^{2}=0$
$\Rightarrow x^{2}-8 x+16+18-16=0$
$\Rightarrow(x-4)^{2}+2=0$
$\Rightarrow(x-4)^{2}=-2$
$\Rightarrow(x-4)=\pm \sqrt{-2}$
But, $\sqrt{-2}$ is not a real number.
Hence, the quadratic equation $x^{2}-8 x+18=0$ has no real solution.
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