Question

Solve Quadratic equation by Factorization :

$x {}^{2} + 5x - (a {}^{2} + a - 6) = 0$

Answer 1

hope your understood

Answer 2

$\quad\quad Method 1$

$\implies \: {x}^{2} + 5x - ( {a}^{2} + a - 6) = 0 \\ \\ \implies {x}^{2} + 5x - ( {a}^{2} + 3a - 2a - 6) = 0 \\ \\ \implies {x}^{2} + 5x - \bigg(a(a + 3) - 2(a + 3) \bigg) = 0 \\ \\ \implies {x}^{2} + 5x - (a + 3)(a - 2) = 0$



Now, on Comparing the formed equation with ax² + bx + c = 0, we get :

a = 1 , b = 5 , c = ( a + 3 )( a + 2 )



We know that in the method of splitting middle term the product of a and c should be equal to the product of two numbers whose sum or difference results b.



In the question,
product of a and c = 1 × ( a + 3 )( a + 2 )
product of a and c = ( a + 3 )( a + 2 ).



sum of two number whose results remains equal to 5 = 3 - 2

But the requirements for splitting middle term are not satisfactory. therefore adding and subtracting a.

sum of two number whose results remains equal to 5 = 3 - 2 + a - a
sum of two number whose results remains equal to 5 = ( a + 3 ) + ( a - 2 )




By splitting middle,


$\implies {x}^{2} +[( a + 3 ) - ( a - 2 ) ]x- ( a + 3 )( a - 2 ) = 0 \\ \\ \implies \: {x}^{2} + (a + 3)x- (a - 2)x- (a + 3)(a - 2) = 0 \\ \\ \implies \: x(x + a + 3) - (a- 2)(x + a + 3)= 0 \\ \\ \implies \: (x + a + 3)(x - a + 2) = 0 \\ \\ \mathsf{ x = - a - 3 \: \: \: \: \: \: or \: \: \: \: \: \: x = a - 2 }$




$\quad \quad Method 2 :$

$\implies \: {x}^{2} + 5x - ( {a}^{2} + a - 6) = 0 \\ \\ \implies \: {x}^{2} + 5x - {a}^{2} - a + 6 = 0 \\ \\ \implies \: x {}^{2} + 5x + 6 = {a}^{2} + a \\ \\ \implies \: {x}^{2} + 2x + 3x + 6 = a(a + 1) \\ \\ \implies x(x + 2) + 3(x + 2) = a(a + 1) \\ \\ \implies \: (x + 3)(x + 2) = a(a + 1) \\ \\ \implies(x + 3)(x + 2) = (a + 0)(a + 1)$


Now, as the difference of 3 and 2 is equal to the difference between 1 and 0. We can take ( x + 3 ) and ( x + 2 ) equal to ( a + 1 ) and ( a + 0 ) respectively.



therefore,
x + 3 = a + 1
x + 2 = a
x = a - 2.


Method 2 is based on things given just above, it can not be applied in all the questions.



Therefore, x = - a - 3 or a - 2.