Question

How do you solve `m^ { 2} + 12m + 71= 0`?

Answer 1

`m=-6+sqrt(35)i,` `-6-sqrt(35)i`

Refer to the explanation for the process.

Explanation:

Solve:

`m^2+12m+71=0` is a quadratic equation in standard form:

`a^2+bx+c=0`,

where:

`a=1`, `b=12`, and `c=71`

Use the quadratic formula to solve for `m`.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Plug in the known values.

`m=(-12+-sqrt((12)^2-4*1*71))/(2*1)`

Simplify.

`m=(-12+-sqrt(-140))/2`

Prime factorize `140`.

`m=(-12+-sqrt(-2xx2xx5xx7))/2`

Simplify.

`m=(-12+-2sqrt(35)i)/2`

Reduce.

`m=(-color(red)cancel(color(black)(12))^6+-color(red)cancel(color(black)(2))^1sqrt(35)i)/color(red)cancel(color(black)(2))^1`

`m=-6+-sqrt(35)i`

Solutions for `m`.

`m=-6+sqrt(35)i,` `-6-sqrt(35)i`

Answer 2

there is no solution for any real number m to solve the equation.

Explanation:

we usually we can factor out the equation make it look like (a-x)(b-x)=0, the solve for the two x's.
But since it doesn't seem as an obvious whole number, we can resort to solving it via the quadratic formula:
`(-b+-sqrt(b^2-4ac))/(2a)`

the general expression is:
`ax^2+bx+c=0`
we have:
`m^2+12m+71=0`
so: a=1, b=12, c=71

now plug in the formula,
`m_1=(-12-sqrt(12^2-4*1*71))/(2*1)`
`m_2=(-12+sqrt(12^2-4*1*71))/(2*1)`

the problem here, is that we know for all real numbers, anything under a square root, or a root in general must be equal or greater than zero, but if we take a closer look to what's under the root, we find it -140
`12^2-4*1*71=-140`
which would result into an error on your calculator, or would show up an answer (if it had that button to solve directly) two answers which are: 0.85+3.06i and 0.85-3.06i. These are complex numbers, that are not contained within the real numbers.
So if you are solving in the reals (m is a real number) then, there is no solution for the equation.

if you are solving in the complex numbers, which i assume you are studying a unit about, then those are your two answers, message me if you want details.