Answer:
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Step-by-step explanation:
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Let us first split the middle term 3x as 2x + 1x [because (2x) × (1x) = 2x² = x² × 2]
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[tex]\dashrightarrow \sf \: \: \: x^2 + 3x + 2 = 0 \\ \\ [/tex]
[tex]\dashrightarrow \sf \: \: \: x^2 + 2x + 1x + 2 \\ \\ [/tex]
[tex]\dashrightarrow \sf \: \: \: x(x + 2) +1 (x + 2) \\ \\ [/tex]
[tex]\dashrightarrow \sf \: \: \: (x + 1)(x + 2) \\ \\ [/tex]
- x² + 3x +2= 0 can be rewritten as (x + 1)(x + 2) = 0
[tex]\: [/tex]
- So, the values of x for which x² + 3x +2= 0 are the same for which (x + 1)(x + 2)
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i.e., either x + 1 = 0 or x + 2 = 0.
[tex]\: [/tex]
Now,
[tex]\dashrightarrow \sf \: \: \: x + 1 = 0 \\ \\ [/tex]
[tex] \dashrightarrow \sf \: \: \:
x = -1 \\ \\ [/tex]
[tex]\dashrightarrow \sf \: \: \: x + 2 = 0 \\ \\ [/tex]
[tex]\dashrightarrow \sf \: \: \: x = -2.
\\ \\ [/tex]
Hence,
[tex]\dashrightarrow \sf \: \: \: x = -1 \: or \: -2 \\ \\ [/tex]
