Question

How many real zeros will x^2+19=8x+3

Answer 1

The solution to the quadratic equation is:

`x=4`

Thus, there will be one real zero.

Explanation:

Given the equation

`x^2+19=8x+3`

Subtract 19 from both sides

`x^2+19-19=8x+3-19`

Simplify

`x^2=8x-16`

Subtract 8x from both sides

`x^2-8x=8x-16-8x`

Simplify

`x^2-8x=-16`

Add (-4)² to both sides

`x^2-8x+\left(-4\right)^2=-16+\left(-4\right)^2`

`x^2-8x+\left(-4\right)^2=0`

Apply perfect square rule: (a-b)² = a² - 2ab + b²

`\left(x-4\right)^2=0` ∵
`x^2-8x+\left(-4\right)^2=\left(x-4\right)^2`

so solve

`x-4=0`

Add 4 to both sides

`x-4+4=0+4`

Simplifying

`x=4`

Therefore, the solution to the quadratic equation is:

`x=4`

Thus, there will be one real zero.