State whether the quadratic equation $(x+4)^{2}-8x=0$ has two distinct real roots. Justify your answer.
(NO) Given equation is $(x+4)^{2}-8x=0$.
Expanding the term: $x^{2}+16+8x-8x=0$ (using $(a+b)^{2}=a^{2}+2ab+b^{2}$).
Simplifying: $x^{2}+16=0$.
Writing in standard form $ax^{2}+bx+c=0$: $x^{2}+0x+16=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=1, b=0, c=16$.
The discriminant $D$ is given by $D=b^{2}-4ac$.
Substituting the values: $D=(0)^{2}-4(1)(16) = 0-64 = -64$.
Since $D < 0$,the quadratic equation has no real roots (it has two imaginary roots).
Expanding the term: $x^{2}+16+8x-8x=0$ (using $(a+b)^{2}=a^{2}+2ab+b^{2}$).
Simplifying: $x^{2}+16=0$.
Writing in standard form $ax^{2}+bx+c=0$: $x^{2}+0x+16=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=1, b=0, c=16$.
The discriminant $D$ is given by $D=b^{2}-4ac$.
Substituting the values: $D=(0)^{2}-4(1)(16) = 0-64 = -64$.
Since $D < 0$,the quadratic equation has no real roots (it has two imaginary roots).
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