Question

find the roots of quadratic equation x2 + 5x - (a+1) (a+6) =0 , where a is constant​

Answer 1

Answer:

$\boxed{\bf \: x = - (a + 6) \: , \: a + 1 \: }$

Step-by-step explanation:

Given quadratic equation is

$\sf \: {x}^{2} + 5x - (a + 1)(a + 6) = 0$

$\sf \: {x}^{2} + (6 - 1)x - (a + 1)(a + 6) = 0$

$\sf \: {x}^{2} + (6 - 1 + a - a)x - (a + 1)(a + 6) = 0$

$\sf \: {x}^{2} + [ (a + 6) - (a + 1)]x - (a + 1)(a + 6) = 0$

$\sf \: {x}^{2} + (a + 6)x - (a + 1)x - (a + 1)(a + 6) = 0$

$\sf \: x[ x + (a + 6)] - (a + 1)[ x + (a + 6)] = 0$

$\sf \: [ x + (a + 6)] \: [ x - (a + 1)] = 0$

$\sf \: x + (a + 6) = 0 \: \:or \: \: x - (a + 1)= 0$

$\implies\sf \: x = - (a + 6) \: \: or \: \: x = a + 1$

Hence,

$\implies\sf \:\boxed{\bf \: x = - (a + 6) \: , \: a + 1 \: }$

$\rule{190pt}{2pt}$

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

Answer 2

Step-by-step explanation:

x² + 5x − (a+1) (a+6) = 0

x² + (6−1) x − (a+1) (a+6) = 0

x² + (6−1 + ba − a) x − (a + 1) (a + 6) = 0

x² + [(a + 6) − (a + 1)] x − (a+1) (a+6) = 0

x² +(a + 6)x − (a + 1) x −(a + 1)(a + 6) = 0

x[x + (a + 6)] − (a + 1)[x + (a + 6)] = 0

[x + (a + 6)] [x − (a + 1)] = 0

x + (a + 6) = 0 or x −(a+1)

x = −(a+6) or x = a + 1

$\color{aqua}{ \rule{500pt}{2pt}}$

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