Question

find the values of a and b such that x^2+3x+4=(x+a)^2+b​

Answer 1

Answer:

a = $\sqrt{x^2+3x+4-b} -x$, $-\sqrt{x^2+3x+4-b}-x$

b = $x^{2} +3x+4-(x+a)^2$

Step-by-step explanation:

Answer 2

Answer:

a= 1.5 ; b = 7/4

Step-by-step explanation:

Magic...

in fact, the key here is

(x+a)^2 = x^2 + 2×x×a +a^2

I believe you can take it from here?

also my calculation can be wrong I did in my head cuz I dont have a proper calculator