(x² + 3x + 2 )² -8(x²+3x)-4=0
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(x² + 3x + 2 )² -8(x²+3x)-4=0
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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
[tex]\rm\bold{Correct ~Question :}[/tex]
[tex]\begin{gathered}\normalsize\bf\ (x^2 + 3x + 2)2 - 8(x^2 - 3x) - 4 = 0 \\ \normalsize\sf\ Find \: the \: value \: of \: x \end{gathered}[/tex]
[tex]\rm\bold{AnswEr :}[/tex]
[tex]☯ \underline{\textsf{According \: to \: the \: given \: in \: Question:}} [/tex]
[tex]\begin{gathered}\normalsize\:\sf\ Break \: the \: question \: into \: two \: parts \: \\ \normalsize\sf\ For \: ex - \end{gathered} [/tex]
[tex]\boxed{\normalsize\sf\underbrace{(x^2 + 3x + 2)2}_{Part \: 1} - \underbrace{8(x^2 - 3x) - 4}_{Part \: 2} = 0} [/tex]
[tex]\underline{\bigstar\:\textsf{Solving \: with \: part \: 1}} [/tex]
[tex]\normalsize\ : \implies\sf\ (x^2 - 3x + 2)2 = 0 [/tex]
[tex]\scriptsize\tt{\quad\dag\ Multiply \: equation \: 1 \: with \: 2}[/tex]
[tex]\normalsize\ : \implies\sf\ 2x^2 - 6x + 4 = 0 [/tex]
[tex]\normalsize\ : \implies\sf\ 2x^2 - 6x + 4 = 0 \: ---(\frak\ Eq.1)[/tex]
[tex]\underline{\bigstar\:\textsf{Solving \: with \: part \: 2}} [/tex]
[tex]\normalsize\twoheadrightarrow\sf\ 8(x^2 + 3x) - 4 = 0[/tex]
[tex]\normalsize\twoheadrightarrow\sf\ 8x^2 - 24x - 4 = 0[/tex]
[tex]\normalsize\twoheadrightarrow\sf\ 8x^2 - 24x - 4 = 0 --- (\frak\ Eq.2)[/tex]
[tex]\underline{\bigstar\:\textsf{Subtract\: Equation \: 1 \: from \: 2(Eq. 1 - Eq.2)}} [/tex]
[tex]\normalsize\dashrightarrow\sf\ [2x^2 + 6x +4] - [8x^2 - 24x - 4] = 0[/tex]
[tex]\normalsize\dashrightarrow\sf\ 2x^2 + 6x + \cancel{4} - 8x^2 - 24x - \cancel{4} = 0[/tex]
[tex]\normalsize\dashrightarrow\sf\ -6x^2 - 18x = 0[/tex]
[tex]\normalsize\dashrightarrow\sf\ -(6x^2 + 18x) = 0[/tex]
[tex]\scriptsize\tt{\quad\dag\ Take \: negative \: sign \: common}[/tex]
[tex]\normalsize\dashrightarrow\sf\ 6x(x + 3) = 0[/tex]
[tex]\begin{gathered}\sf{The \: Value \: of \: x \: }\begin{cases}\sf{6x = 0}\\\sf{(x + 3) = 0}\end{cases}\end{gathered} [/tex]
[tex]\normalsize\ : \implies\sf\ 6x \quad\ or \quad\ (x + 3) = 0 [/tex]
[tex]\normalsize\ : \implies\sf\ x = \frac{0}{6} \quad\ or \quad\ 0 - 3 [/tex]
[tex]\normalsize\ : \implies\sf\ x = 0 \quad\ or \quad\ -3[/tex]
[tex]\therefore\:\underline{\textsf{Hence, \: the \: value \: of \: x \: are }{\textbf{\: 0, -3}}}[/tex]