Question

find the roots of the quadratic equation x2+5x-(alpha+1)(alpha+6)=0,:where:alpha is some constant... please help its urgent..

Answer 1

Answer:

α + 1, -(α + 6)

Step-by-step explanation:

Given Quadratic equation is x² + 5x - (α + 1)(α + 6) = 0

⇒ x² + 6x - x - (α + 1)(α + 6) = 0

⇒ x² + (α + 6)x - (α + 1)x - (α + 1)(α + 6) = 0

⇒ x[x + (α + 6)] - (α + 1)[x + (α + 6)] = 0

⇒ [x - (α + 1)][x + (α + 6)] = 0

⇒ [x - (α + 1)] = 0, [x + (α + 6)] = 0

⇒ x = α + 1, -(α + 6).

Therefore, the roots of the quadratic equation is : (α + 1), -(α + 6).

Hope it helps!

Answer 2

Answer:

[Tex] x = \alpha + 1 [/Tex]

[Tex] x = -\alpha -6 [/Tex]

Step-by-step explanation:

[Tex] x^2 + 5x - (\alpha + 1)(\alpha +6) = 0 [/Tex]

[Tex] x^2 + 5x - \alpha^2 - 7\alpha - 6= 0 [/Tex]

[Tex] x^2 + 5x = \alpha^2 + 7\alpha +6 [/Tex]

[Tex] x^2 + 5x + 25/4 = \alpha^2 + 7\alpha +6 + 25/4 [/Tex]

[Tex] (x + 5/2)^2 = \alpha^2 + 7\alpha +6 + 25/4 [/Tex]

[Tex] (x + 5/2)^2 = \alpha^2 + 7\alpha + 49/4 [/Tex]

[Tex] (x + 5/2)^2 = (\alpha + 7/2)^2 [/Tex]

[Tex] x + 5/2 = +/- (\alpha + 7/2) [/Tex]

[Tex] x = \alpha + 1 [/Tex]

[Tex] x = -\alpha -6 [/Tex]