Question

x^2+3x-18 split the middle term​

Answer 1

Question :-

Split the middle term of-

$\to \rm {x}^{2} + 3x - 18 = 0$

Answer :-

$\to \: \: \boxed{ \rm \: x = - 6 \: \: \: \: or \: \: \: \: 3}$

Solution :-

We have ,

$\to \rm \: {x}^{2} + 3x - 18 = 0 \\ \\ \bf \rm split \: the \: middle \: term \\ \\ \to \rm \: {x}^{2} + (6 - 3)x - 18 = 0 \\ \\ \to \rm \: {x}^{2} + 6x - 3x - 18 = 0 \\ \\ \to \rm \: x(x + 6) - 3(x + 6) = 0 \\ \\ \to \rm \: \: (x + 6)(x - 3) = 0 \\ \\ \: \star \bf \: case \: (1) \: if \: \\ \to \rm \: x + 6 = 0 \\ \: \\ \to \boxed{ \rm \: x = - 6} \\ \\ \star \: \bf \: case \: (2) \: if \: \\ \to \rm \: x - 3 = 0 \\ \\ \to \boxed{ \rm \: x = 3} \\ \\ \sf \: hence \rm \: x = 3 \: or \: - 6 \\ \\ \underline{ \underline{ \sf \red{ \sf \: \boxed{ \sf \: verification \: }}}} : - \\ \\ \rm \: firstly \: put \: x = - 6 \\ \\ \to \rm {( - 6)}^{2} + 3 \times ( - 6) - 18 = 0 \\ \\ \to \: 36 - 18 - 18 = 0 \\ \\ \to \: 0 = 0 \\ \\ \rm \: now \: put \: x = 3 \\ \\ \to \rm {(3)}^{2} + 3 \times 3 - 18 = 0 \\ \\ \to \: 9 + 9 - 18 = 0 \\ \\ \to \: 0 = 0 \\ \\ \red{\bf hence \: verified}$