Question

(x² + 3x + 2)2 – 8(x² + 3x) – 4 = 0
​

Answer 1

Answer:

Step-by-step explanation:

2(x^2 + 3x +2) - 8(x^2 + 3x) - 4 = 0

=> 2x^2 + 6x + 4 - 8x^2 - 24x - 4 = 0

=> -6x^2 - 18x = 0

=> 6x^2 + 18x = 0

=> 6(x^2 + 3x) = 0

=> x^2 + 3x = 0

=> x(x + 3) = 0

=> x + 3 = 0

Therefore, x = -3..........

Thank You...

Answer 2

Correct Question :

$\normalsize\bf\ (x^2 + 3x + 2)2 - 8(x^2 - 3x) - 4 = 0 \\ \normalsize\sf\ Find \: the \: value \: of \: x$

$\rule{200}2$

AnswEr :

☯ $\underline{\textsf{According \: to \: the \: given \: in \: Question:}}$

$\normalsize\:\sf\ Break \: the \: question \: into \: two \: parts \: \\ \normalsize\sf\ For \: ex -$

$\boxed{\normalsize\sf\underbrace{(x^2 + 3x + 2)2}_{Part \: 1} - \underbrace{8(x^2 - 3x) - 4}_{Part \: 2} = 0}$

$\rule{170}1$

$\underline{\bigstar\:\textsf{Solving \: with \: part \: 1}}$

$\normalsize\ : \implies\sf\ (x^2 - 3x + 2)2 = 0$

$\scriptsize\tt{\quad\dag\ Multiply \: equation \: 1 \: with \: 2}$

$\normalsize\ : \implies\sf\ 2x^2 - 6x + 4 = 0$

$\normalsize\ : \implies\sf\ 2x^2 - 6x + 4 = 0 \: ---(\frak\ Eq.1)$

$\underline{\bigstar\:\textsf{Solving \: with \: part \: 2}}$

$\normalsize\twoheadrightarrow\sf\ 8(x^2 + 3x) - 4 = 0$

$\normalsize\twoheadrightarrow\sf\ 8x^2 - 24x - 4 = 0$

$\normalsize\twoheadrightarrow\sf\ 8x^2 - 24x - 4 = 0 --- (\frak\ Eq.2)$

$\rule{170}1$

$\underline{\bigstar\:\textsf{Subtract\: Equation \: 1 \: from \: 2(Eq. 1 - Eq.2)}}$

$\normalsize\dashrightarrow\sf\ [2x^2 + 6x +4] - [8x^2 - 24x - 4] = 0$

$\normalsize\dashrightarrow\sf\ 2x^2 + 6x + \cancel{4} - 8x^2 - 24x - \cancel{4} = 0$

$\normalsize\dashrightarrow\sf\ -6x^2 - 18x = 0$

$\normalsize\dashrightarrow\sf\ -(6x^2 + 18x) = 0$

$\scriptsize\tt{\quad\dag\ Take \: negative \: sign \: common}$

$\normalsize\dashrightarrow\sf\ 6x(x + 3) = 0$

$\rule{170}1$

$\sf{The \: Value \: of \: x \: }\begin{cases}\sf{6x = 0}\\\sf{(x + 3) = 0}\end{cases}$

$\normalsize\ : \implies\sf\ 6x \quad\ or \quad\ (x + 3) = 0$

$\normalsize\ : \implies\sf\ x = \frac{0}{6} \quad\ or \quad\ 0 - 3$

$\normalsize\ : \implies\sf\ x = 0 \quad\ or \quad\ -3$

$\therefore\:\underline{\textsf{Hence, \: the \: value \: of \: x \: are }{\textbf{\: 0, -3}}}$