Question

(x² + 6x + 1) = (x – 3)

Answer 1

Hello!

$\large\boxed{x = -4, -1}$

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

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf\implies x^2 + 6x + 1 = x - 3$

$\sf\implies x^2 + 6x - x + 1 + 3 = 0$

⠀⠀⠀⠀

$\sf\implies x^2 + 5x + 4 = 0$

⠀⠀⠀⠀⠀⠀⠀

compare the eq with $\sf{\underline{\bold{ax^2 + bx + c = 0 }}}$

⠀⠀⠀⠀⠀⠀⠀

☯ a = 1

☯ b = 5

☯ c = 4

⠀⠀⠀⠀⠀⠀⠀

now :-

⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\pink{\large{\mathfrak{x = \dfrac{ - b \pm \sqrt D }{2a }}}}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\pink{\large{\mathfrak{ D = b^2 - 4ac }}}}}}$

⠀⠀⠀⠀⠀⠀⠀

finding value of D.

⠀⠀⠀⠀⠀⠀⠀

$\sf\implies D = b^2 - 4ac$

$\sf\implies D = (5)^2 - 4 \times 1 \times 4$

$\sf\implies D = 25 - 16$

⠀⠀⠀⠀

$\sf\implies D = 9$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 9}}}}}}$

⠀⠀⠀⠀⠀⠀⠀

putting values in the eq.

⠀⠀⠀⠀⠀⠀⠀

$\sf\implies x = \dfrac{ -b \pm\sqrt D }{2a}$

$\sf\implies x = \dfrac{ -( 5) \pm\sqrt {9} }{2\times 1 }$

$\sf\implies x = \dfrac{ -5 \pm 3 }{2}$

⠀⠀⠀⠀⠀⠀⠀

✒$\sf x = \dfrac{ -5 + 3 }{ 2 }$

$\implies x = \dfrac {-2}{2}$

$\implies x = -1$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = -1 }}}}}}$

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀

✒$\sf x = \dfrac{ -5 - 3 }{ 2 }$

$\implies x = \dfrac {-8}{2}$

$\implies x = -4$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = -4 }}}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\purple{\large{\bold{ x = -1 \: or \:-4 }}}}}}$