Answer:
The solution to the quadratic equation is:
[tex]x=4[/tex]
Thus, there will be one real zero.
Step-by-step explanation:
Given the equation
[tex]x^2+19=8x+3[/tex]
Subtract 19 from both sides
[tex]x^2+19-19=8x+3-19[/tex]
Simplify
[tex]x^2=8x-16[/tex]
Subtract 8x from both sides
[tex]x^2-8x=8x-16-8x[/tex]
Simplify
[tex]x^2-8x=-16[/tex]
Add (-4)² to both sides
[tex]x^2-8x+\left(-4\right)^2=-16+\left(-4\right)^2[/tex]
[tex]x^2-8x+\left(-4\right)^2=0[/tex]
Apply perfect square rule: (a-b)² = a² - 2ab + b²
[tex]\left(x-4\right)^2=0[/tex] ∵ [tex]x^2-8x+\left(-4\right)^2=\left(x-4\right)^2[/tex]
so solve
[tex]x-4=0[/tex]
Add 4 to both sides
[tex]x-4+4=0+4[/tex]
Simplifying
[tex]x=4[/tex]
Therefore, the solution to the quadratic equation is:
[tex]x=4[/tex]
Thus, there will be one real zero.
