Hello!
[tex]\large\boxed{x = -4, -1}[/tex]
x² + 6x + 1 = x - 3
Bring all terms to one side by subtracting x and adding 3:
x² + 6x - x + 1 + 3 = x - x - 3 + 3
x² + 5x + 4 = 0
Factor:
(x + 4)(x + 1) = 0
Set each factor to 0:
x + 4 = 0
x = -4
x + 1 = 0
x = -1
[tex]\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}[/tex]
[tex]\sf\implies x^2 + 6x + 1 = x - 3 [/tex]
[tex]\sf\implies x^2 + 6x - x + 1 + 3 = 0[/tex]
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[tex]\sf\implies x^2 + 5x + 4 = 0 [/tex]
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compare the eq with [tex]\sf{\underline{\bold{ax^2 + bx + c = 0 }}}[/tex]
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☯ a = 1
☯ b = 5
☯ c = 4
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now :-
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[tex]\sf{\underline{\boxed{\pink{\large{\mathfrak{x = \dfrac{ - b \pm \sqrt D }{2a }}}}}}}[/tex]
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[tex]\sf{\underline{\boxed{\pink{\large{\mathfrak{ D = b^2 - 4ac }}}}}}[/tex]
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finding value of D.
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[tex]\sf\implies D = b^2 - 4ac [/tex]
[tex]\sf\implies D = (5)^2 - 4 \times 1 \times 4[/tex]
[tex]\sf\implies D = 25 - 16 [/tex]
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[tex]\sf\implies D = 9 [/tex]
[tex]\sf{\underline{\boxed{\blue{\large{\bold{ D = 9}}}}}}[/tex]
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putting values in the eq.
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[tex]\sf\implies x = \dfrac{ -b \pm\sqrt D }{2a}[/tex]
[tex]\sf\implies x = \dfrac{ -( 5) \pm\sqrt {9} }{2\times 1 }[/tex]
[tex]\sf\implies x = \dfrac{ -5 \pm 3 }{2}[/tex]
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✒[tex] \sf x = \dfrac{ -5 + 3 }{ 2 }[/tex]
[tex]\implies x = \dfrac {-2}{2}[/tex]
[tex]\implies x = -1 [/tex]
[tex]\sf{\underline{\boxed{\purple{\large{\bold{ x = -1 }}}}}}[/tex]
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✒[tex] \sf x = \dfrac{ -5 - 3 }{ 2 }[/tex]
[tex]\implies x = \dfrac {-8}{2}[/tex]
[tex]\implies x = -4 [/tex]
[tex]\sf{\underline{\boxed{\purple{\large{\bold{ x = -4 }}}}}}[/tex]
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[tex]\sf{\underline{\boxed{\purple{\large{\bold{ x = -1 \: or \:-4 }}}}}}[/tex]
