Answer:
[tex]\large \boxed{\sf \ \ \ a+b+c=3 \ \ \ }[/tex]
Step-by-step explanation:
Hello,
[tex]4x^2+2x-1=4(x^2+\dfrac{2}{4}x)-1=4[(x+\dfrac{1}{4})^2-\dfrac{1}{4^2}]-1[/tex]
As
[tex](x+\dfrac{1}{4})^2=x^2+\dfrac{2}{4}x+\dfrac{1^2}{4^2}=x^2+\dfrac{2}{4}x+\dfrac{1}{4^2} \ \ So \\\\x^2+\dfrac{2}{4}x=(x+\dfrac{1}{4})^2-\dfrac{1}{4^2}[/tex]
Let 's go back to the first equation
[tex]4x^2+2x-1=4[(x+\dfrac{1}{4})^2-\dfrac{1}{4^2}]-1=4(x+\dfrac{1}{4})^2-\dfrac{1}{4}-1\\\\=4(x+\dfrac{1}{4})^2-\dfrac{1+4}{4}=\boxed{4(x+\dfrac{1}{4})^2-\dfrac{5}{4}}[/tex]
a = 4
[tex]b=\dfrac{1}{4}\\\\c=-\dfrac{5}{4}[/tex]
[tex]a+b+c=4+\dfrac{1}{4}-\dfrac{5}{4}=\dfrac{4*4+1-5}{4}=\dfrac{12}{4}=3[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
