How do you find all the real and complex roots of $x^{2} + 8 x + 25 = 0$ $x^2 + 8x + 25 = 0$ ?

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How do you find all the real and complex roots of $x^{2} + 8 x + 25 = 0$ $x^2 + 8x + 25 = 0$ ?

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Answer 1 · 2016-01-10T12:23:42

$x = - 4 \pm 3 i$ $x=-4+-3i$

Explanation:

Complete the square.

$(x^{2} + 8 x + 16) + 25 - 16 = 0$ $(x^2+8x+16)+25-16=0$

${(x + 4)}^{2} + 9 = 0$ $(x+4)^2+9=0$

${(x + 4)}^{2} = - 9$ $(x+4)^2=-9$

$\sqrt{{(x + 4)}^{2}} = \sqrt{- 9}$ $sqrt((x+4)^2)=sqrt(-9)$

$x + 4 = \pm 3 i$ $x+4=+-3i$

$x = - 4 \pm 3 i$ $x=-4+-3i$

Answer 2 · 2016-01-10T12:27:09

Roots (both complex) are at $x = - 4 + 3 i$ $x=-4+3i$ and $x = - 4 - 3 i$ $x=-4-3i$

Explanation:

The quadratic formula tells us that for a quadratic equation in the form:
$XXX a x^{2} + b x + c = 0$ $color(white)("XXX")ax^2+bx+c=0$
the roots are given by:
$XXX x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)$

For the given example: $x^{2} + 8 x + 25 = 0$ $x^2+8x+25=0$
$XXX a = 1, b = 8, and c = 25$ $color(white)("XXX")a=1, b=8, and c=25$

So the roots are:
$XXX x = \frac{- 8 \pm \sqrt{8^{2} - 4 (1) (25)}}{2 (1)}$ $color(white)("XXX")x=(-8+-sqrt(8^2-4(1)(25)))/(2(1))$

$XXXX = \frac{- 8 \pm \sqrt{- 36}}{2}$ $color(white)("XXXX")=(-8+-sqrt(-36))/2$

$XXXX = \frac{- 8 \pm \sqrt{36} \cdot \sqrt{- 1}}{2}$ $color(white)("XXXX")=(-8+-sqrt(36)*sqrt(-1))/2$

$XXXX = - 4 \pm 3 i$ $color(white)("XXXX")=-4+-3i$

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How do you find all the real and complex roots of $x^{2} + 8 x + 25 = 0$ $x^2 + 8x + 25 = 0$ ?

2 Answers

Explanation:

Explanation:

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How do you find all the real and complex roots of x2+8x+25=0x^2 + 8x + 25 = 0?

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How do you find all the real and complex roots of $x^{2} + 8 x + 25 = 0$ $x^2 + 8x + 25 = 0$ ?