Question

Solve the following question

Solve for x

${x}^{2} + x - (a + 1)(a + 2) = 0$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

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

can be rewritten as

$\rm :\longmapsto\: {x}^{2} + \red{(2 - 1)}x - (a + 1)(a + 2) = 0$

can be further rewritten as

$\rm :\longmapsto\: {x}^{2} + \red{(2 - 1 + a - a)}x - (a + 1)(a + 2) = 0$

$\rm :\longmapsto\: {x}^{2} + \red{[(a + 2) - (a + 1)]}x - (a + 1)(a + 2) = 0$

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

$\rm :\longmapsto\:x[x + (a + 2)] - (a + 1)[x + (a + 2)] = 0$

$\rm :\longmapsto\:[x + (a + 2)][x - (a + 1)] = 0$

$\bf\implies \:x = - (a + 2) \: \: \: or \: \: \: x = a + 1$

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LEARN MORE :-

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

Question:-

x² + x – (a + 1) (a + 2) = 0.

Given:-

Quadratic Equation: x² + x – (a + 1) (a + 2) = 0.

To Find:-

• Needed to solve x from the given Equation.

Solution:-

x² + x – (a + 1) (a + 2) = 0.

=> x² + [2 – 1 + a – a] x – (a + 1) (a + 2) = 0.

=> x² + [ (a + 2) – (a + 1) ] x – (a + 1) (a + 2) = 0.

=> x² + (a + 2) x – (a + 1) x – (a + 1) (a + 2) = 0.

=> x [ x + (a + 2) ] – (a + 1) [ x + (a + 2) ] = 0.

=> [ x + (a + 2) ] [ x – (a + 1) ] = 0.

=> x + (a + 2) = 0. x – (a + 1) = 0.

=> x = – (a + 2). x = (a + 1).

Answer:-

$\sf \large { \sf \red {\therefore \:The \: solutions \: are \: x = - (a + 2) }} \\ \sf \large\red{and \: x = a + 1.}$

Hope you have satisfied. ⚘