Question

How do you solve `11v ^ { 2} - 7v + 19= 9`?

Answer 1

`v = (7 +- sqrt(-391))/22`

Explanation:

`11v^2 - 7v + 19 = 9`

The 1st thing we do is to get `0` on one side of the equation so that we can try to factor it (or use quadratic formula if not possible to factor). So we move the 9 to the other side of the equation.

`11v^2 - 7v + 10 = 0`

This equation is in standard form, or `ax^2 + bx + c = 0`

Now we can try to factor it. We have to find two numbers that multiply up to `a * c` (or `11 * 10`) AND add up to `b`, or `-7`.

So 2 numbers that multiply to `110` and add up to `-7`. There's NO number that does so!

So in this case we have to use the quadratic formula.
The quadratic formula equation is:
`v = (-b+- sqrt(b^2 - 4ac))/(2a)`
We know that `a = 11`, `b = -7`, and `c = 10`, so let's put these numbers in the formula.

`v = (-(-7) +- sqrt((-7)^2 - 4(11)(10)))/(2(11))`
`v = (7 +- sqrt(49 - 4(110)))/22`
`v = (7 +- sqrt(49 - 440))/22`
`v = (7 +- sqrt(-391))/22`

Answer 2

x = `(7 + sqrt(489))/(22)` , x = `(7 - sqrt(489))/(22)`

Explanation:

Given:
`11v^2 - 7v + 19 = 9`
Next, we subtract the coefficients;
`11v^2 - 7v + 19 - 9 = 0`
`11v^2 - 7v + 10 = 0`

We proceed to solve this question using the formula:

`ax^2 + bx + c = 0`

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

Substituting the values of a,b and c from `11v^2 - 7v +10 = 0`;
`(-(-7) +- sqrt((-7)^2 - 4*(11)*(10)))/(2*11)`
`((7) +- sqrt(-391))/(22)`

Therefore, the values of x are:
x = `(7 + sqrt(391))/(22)i`
x = `(7 - sqrt(391))/(22)i`

Where i denotes a complex number.