The roots of the quadratic equation x^{2}+18x | vedclass

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(D) The given quadratic equation is $x^{2}+18x+81=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we have $a=1$,$b=18$,and $c=81$.The discrim

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(D) The given quadratic equation is $x^{2}+18x+81=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we have $a=1$,$b=18$,and $c=81$.The discriminant $D$ is given by $D = b^{2}-4ac$.Substituting the values: $D = (18)^{2} - 4(1)(81) = 324 - 324 = 0$.Since the discriminant $D=0$,the quadratic equation has two real and equal roots.Alternatively,we can factorize the equation: $x^{2}+18x+81 = (x+9)^{2} = 0$.This gives $(x+9)(x+9)=0$,so $x = -9, -9$.Thus,the roots are equal.

The roots of the quadratic equation $x^{2}+18x+81=0$ are ..... .

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