Show that the equation x2 + 6x + 6 = 0 has real roots (show that discr

1.	Show that the equation x2 + 6x + 6 = 0 has real roots (show that discriminant > 0) and solve it.
Answer» Given quadratic equation is x² + 6x + 6 = 0 By comparing given quadratic equation with ax² + bx + c = 0, we get a =1, b = 6 and c = 6. The discriminant of given quadratic equation is D = b² − 4ac = 6² − 4 × 1 × 6 = 36 – 24 = 12 > 0 (real and distinct roots) Hence, the roots of given quadratic equation are real and distinct. And the roots are \(\frac{-b±\sqrt{D}}{2a}\) = \(\frac{-6±\sqrt{12}}{2\times1}\) = \(\frac{-6±2\sqrt{3}}{2}\) = -3 ± \(\sqrt3\). Hence, the roots (or solution) of given quadratic equation are x = -3 +√3 and x = -3 − √3.

Show that the equation x2 + 6x + 6 = 0 has real roots (show that discriminant > 0) and solve it.

Answer»

Given quadratic equation is x² + 6x + 6 = 0

By comparing given quadratic equation with ax² + bx + c = 0,

we get a =1, b = 6 and c = 6.

The discriminant of given quadratic equation is

D = b² − 4ac = 6² − 4 × 1 × 6 = 36 – 24 = 12 > 0 (real and distinct roots)

Hence, the roots of given quadratic equation are real and distinct.

And the roots are \(\frac{-b±\sqrt{D}}{2a}\) = \(\frac{-6±\sqrt{12}}{2\times1}\) = \(\frac{-6±2\sqrt{3}}{2}\) = -3 ± \(\sqrt3\).

Hence, the roots (or solution) of given quadratic equation are x = -3 +√3 and x = -3 − √3.

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