Question

How do you solve `x^ { 2} + 98= 14x`?

Answer 1

`x = 7+-7i`

Explanation:

Given:

`x^2+98=14x`

Subtracting `14x` from both sides, we get:

`x^2-14x+98 = 0`

This is in standard form:

`ax^2+bx+c = 0`

with `a=1`, `b=-14` and `c=98`.

It has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (color(blue)(-14))^2-4(color(blue)(1))(color(blue)(98)) = 196-392 = -196`

Since `Delta < 0` we can tell that this quadratic has no Real solutions.

Completing the square

We can solve it with complex numbers, using the difference of squares identity:

`A^2-B^2=(A-B)(A+B)`

with `A=(x-7)` and `B=7i`, (where `i^2=-1`) as follows:

`0 = x^2-14x+98`

`color(white)(0) = x^2-14x+7^2+49`

`color(white)(0) = (x-7)^2+7^2`

`color(white)(0) = (x-7)^2-(7i)^2`

`color(white)(0) = ((x-7)-7i)((x-7)+7i)`

`color(white)(0) = (x-7-7i)(x-7+7i)`

Hence: `x = 7+-7i`

Quadratic formula

If you prefer, you can use the quadratic formula:

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

`color(white)(x) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x) = (-(color(blue)(-14))+-sqrt(-196))/(2(color(blue)(1)))`

`color(white)(x) = (14+-14i)/2`

`color(white)(x) = 7+-7i`