Question

1. The roots of the quadratic equation x2 + 5x - (P + 1)(P + 6) = 0, where p is a
constant are
A. (P + 1), (p+6)
B. (p + 1), - (P + 6)
C. - (P + 1), (p+6)
D. - (p + 1), -(p + 6)
choose the correct option​

Answer 1

Question:

What are the roots of the quadratic equation x^2 + 5x - (p+1)(p+6) = 0.

Solution:

We have;

=> x^2 + 5x - (p+1)(p+6) = 0

=> x^2 + [(p+6)-(p+1)]x - (p+1)(p+6) = 0

=> x^2 + (p+6)x - (p+1)x - (p+1)(p+6) = 0

=> x[x + (p+6)] - (p+1)[x + (p+6)] = 0

=> [x + (p+6)][x - (p+1)] = 0

Case(1)

when , [x + (p+6)] = 0

then x = - (p+6)

Case (2)

when , [x - (p+1)] = 0

then , x = (p+1)

Hence,

The roots of the given equation are:

(p+1) and -(p+1)

Thus,

The required answer is : option (B)

Answer 2

Answer:

→ (P+1) , -(P+6 )

Option B is correct.

$\bf{\underline{ \: step - by - step \: explanation}} \\ \\ \bf{ given \: quadratic \: equation} →\: \\ \\ \\ \implies \: \bf{{x}^{2} + 5x - (P + 1)(P + 6)} = 0 \\ \\ \: → \bf{ {x}^{2} + \big((P + 6) - ( P + 1) \big)x - (P + 1)(P + 6) = 0} \\ \\ →\bf{ {x}^{2} + (P + 6)x - x(P + 1) - (P + 1)(P + 6)} = 0 \\ \\ → \bf{x \big(x + (P + 6) \big) - (P + 1) \big(x + (P + 6)} = 0 \\ \\ →\bf{ \big(x -(P + 1) \big) \big(x + (P + 6) \big) = 0} \\ \\ \bf{\star \: if \: \: \: \big(x - (P + 1) \big) = 0} \\ \therefore \: \bf{x = ( P + 1) }\\ \\ \: \star \bf{ if \: \: \: \big(x + (P + 6) \big)= 0} \\ \therefore \: \bf{x = - (P + 6)} \:$

Option B is correct.