How do you solve $x^{2} + 4 x + 7 = 0$ $x^2+4x+7=0$ ?

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Answer 1 · 2015-08-12T14:45:32

Try factoring first. It won't work, so either use the quadratic formula or complete the square, as you prefer.

$x^{2} + 4 x + 7 = 0$ $x^2+4x+7=0$

The expression will not factor using integers.

Use your favorite of either the quadratic formula or completing the square.

Completing the square

Because the leading coefficient is $1$ $1$ , and the coefficient of $x$ $x$ is even, completing the square is a good choice for this equation.

$x^{2} + 4 x = - 7$ $x^2+4x " " " " = -7$

$x^{2} + 4 x + 4 = - 7 + 4$ $x^2+4x +4 = -7+4$

${(x + 2)}^{2} = - 3$ $(x+2)^2 = -3$

Which has no solution in the Real Numbers, but has 2 imaginary solutions in the Complex Numbers.

$x + 2 = \pm \sqrt{- 3} = \pm i \sqrt{3}$ $x+2 = +-sqrt(-3) = +-isqrt3$

$x = - 2 \pm i \sqrt{3}$ $x = -2 +- i sqrt3$

Quadratic formula
$x^{2} + 4 x + 7 = 0$ $x^2+4x+7=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- (4) \pm \sqrt{{(4)}^{2} - 4 (1) (7)}}{2 (1)}$ $x = (-(4)+-sqrt((4)^2-4(1)(7)))/(2(1))$

$= \frac{- 4 \pm \sqrt{16 - 28}}{2}$ $= (-4+-sqrt(16-28))/2$

$= \frac{- 4 \pm \sqrt{- 12}}{2}$ $= (-4+-sqrt(-12))/2$

Which has no solution in the Real Numbers, but has 2 imaginary solutions in the Complex Numbers.

$= \frac{- 4 \pm 2 i \sqrt{3}}{2}$ $= (-4+- 2isqrt(3))/2$

$= - 2 \pm i \sqrt{3}$ $= -2 +- isqrt3$

How do you solve x2+4x+7=0x^2+4x+7=0?