Question

I need help understanding quadratic equations with a not equal to 1

Answer 1

we have

f(x)=4x^2+8x+1

Complete the square

step 1

Factor the leading coefficient 4

f(x)=4(x^2+2x)+1

step 2

f(x)=4(x^2+2x+1-1)+1

f(x)=4(x^2+2x+1)+1-4

f(x)=4(x^2+2x+1)-3

step 3

Rewrite as perfect square

f(x)=4(x+1)^2-3

step 4

Determine the solutions

equate the function f(x) to zero

4(x+1)^2-3=0

4(x+1)^2=3

(x+1)^2=3/4

square root both sides

`\begin{gathered} x+1=\pm\frac{\sqrt[]{3}}{2} \\ \\ x=-1\pm\frac{\sqrt[]{3}}{2} \end{gathered}`

step 2

we have

f(x)=4(x^2+2x)+1

the number 2 divided by 2 is 1

so

add and subtract 1 not change the equation

f(x)=4(x^2+2x+1-1)+1

Apply the distributive property to -1

f(x)=4(x^2+2x+1)+4(-1)+1

f(x)=4(x^2+2x+1)-4+1

f(x)=4(x^2+2x+1)-3