the roots of the equation x^2+x-p(p+1)=0 where p is a constant are (1) p,p+1 (2) p,-(p+1)=0 (3) -p,-(p=+1)
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the roots of the equation x^2+x-p(p+1)=0 where p is a constant are (1) p,p+1 (2) p,-(p+1)=0 (3) -p,-(p=+1)
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Answer:
p, -(p+1)
Step-by-step explanation:
Using discrimination formula u can easily find answers
Verified answer
Answer:
[tex] \qquad\qquad\boxed{ \sf{ \: - ( p + 1), \: \: \: \: p \: }} \\ \\ [/tex]
Step-by-step explanation:
Given quadratic equation is
[tex]\sf \: {x}^{2} + x - p(p + 1) = 0 \\ \\ [/tex]
can be rewritten as
[tex]\sf \: {x}^{2} + (p + 1 - p)x - p(p + 1) = 0 \\ \\ [/tex]
[tex]\sf \: {x}^{2} + (p + 1)x - px - p(p + 1) = 0 \\ \\ [/tex]
[tex]\sf \: x(x + p + 1) - p(x + p + 1) = 0 \\ \\ [/tex]
[tex]\sf \: (x + p + 1) \: (x - p) = 0 \\ \\ [/tex]
[tex]\sf\implies x + p + 1 = 0 \: \: or \: \: x - p = 0 \\ \\ [/tex]
[tex]\sf\implies x = - ( p + 1) \: \: or \: \: x = p \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
Nature of roots :-
Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.
If Discriminant, D > 0, then roots of the equation are real and unequal.
If Discriminant, D = 0, then roots of the equation are real and equal.
If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.
Where,
Discriminant, D = b² - 4ac