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Solve x^2−6x+15=0 by completing the square

User Mossaab
by
5.8k points

1 Answer

4 votes

Completing the square is represented by:


x^2+bx+((b)/(2))^2+c=d+((b)/(2))^2

Then,


\begin{gathered} x^2-6x+((-6)/(2))^2=d+(-(6)/(2))^2 \\ x^2-6x+9+15=0+9 \\ x^2-6x+24=9 \\ \end{gathered}

By the rule of discriminant, we know that this expression has not real solutions.


\text{Discriminant}=(b^2-4ac)

If the discriminant is negative, the quadratic equation doesn't have real solutions.

User Nikolaj Dam Larsen
by
5.2k points
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