Show that the quadratic equation
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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