Question

(x² + 3x + 2 )² -8(x²+3x)-4=0​

Answer 1

Step-by-step explanation:

Given

An equation

(x² + 3x)² – (x² + 3x) –6 = 0.

{\sf{\green{\underline{\large{To\:find}}}}}

Tofind

Roots of the equation

{\sf{\pink{\underline{\Large{Explanation}}}}}

Explanation

Let x²+3x as y

Reasons

Simplify by the equation

easily find the roots.

New equation

(x² + 3x)² – (x² + 3x) –6 = 0.

=>y²-y-6=0

Spilt the middle term in such a way that the product will be 6 and sum will be-1

=>y²-3y+2y-6=0

=>y(y-3)+2(y-3)=0

=>(y-3) (y+2)=0

=>y=3,-2

Taking y=3

X²+3x=-2

=>x²+3x+2

=>x²+2x+x+2

=>x(x+2)+1(x+2)

=>(x+1)(x+2)

=>x=-1,-2.

Hence roots if equation be -1 and -2

Answer 2

$\rm\bold{Correct ~Question :}$

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

$\rm\bold{AnswEr :}$

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

$\begin{gathered}\normalsize\:\sf\ Break \: the \: question \: into \: two \: parts \: \\ \normalsize\sf\ For \: ex - \end{gathered}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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