How do you solve 11v ^ { 2} - 7v + 19= 911v27v+19=9?

2 Answers
Nov 23, 2017

v = (7 +- sqrt(-391))/22v=7±39122

Explanation:

11v^2 - 7v + 19 = 911v27v+19=9

The 1st thing we do is to get 00 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 = 011v27v+10=0

This equation is in standard form, or ax^2 + bx + c = 0ax2+bx+c=0

Now we can try to factor it. We have to find two numbers that multiply up to a * cac (or 11 * 101110) AND add up to bb, or -77.

So 2 numbers that multiply to 110110 and add up to -77. 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)v=b±b24ac2a
We know that a = 11a=11, b = -7b=7, and c = 10c=10, so let's put these numbers in the formula.

v = (-(-7) +- sqrt((-7)^2 - 4(11)(10)))/(2(11))v=(7)±(7)24(11)(10)2(11)
v = (7 +- sqrt(49 - 4(110)))/22v=7±494(110)22
v = (7 +- sqrt(49 - 440))/22v=7±4944022
v = (7 +- sqrt(-391))/22v=7±39122

Nov 23, 2017

x = (7 + sqrt(489))/(22)7+48922 , x = (7 - sqrt(489))/(22)748922

Explanation:

Given:
11v^2 - 7v + 19 = 911v27v+19=9
Next, we subtract the coefficients;
11v^2 - 7v + 19 - 9 = 011v27v+199=0
11v^2 - 7v + 10 = 011v27v+10=0

We proceed to solve this question using the formula:

ax^2 + bx + c = 0ax2+bx+c=0

(-b +- sqrt(b^2 - 4ac))/(2a)b±b24ac2a

Substituting the values of a,b and c from 11v^2 - 7v +10 = 011v27v+10=0;
(-(-7) +- sqrt((-7)^2 - 4*(11)*(10)))/(2*11)(7)±(7)24(11)(10)211
((7) +- sqrt(-391))/(22)(7)±39122

Therefore, the values of x are:
x = (7 + sqrt(391))/(22)i7+39122i
x = (7 - sqrt(391))/(22)i739122i

Where i denotes a complex number.