Question

X^2 + 4x - 9 = (x + a)^2 + b Find the value of a and the value of b

Answer 1

Answer:

a = 2, b = -13

Step-by-step explanation:

We have three ways.

$(a+b)^2=a^2+2ab+b^2\qquad(*)$

$\bold{1.}\\\\x^2+4x-9=x^2+2(x)(2)-9=\underbrace{x^2+2(x)(2)+2^2}_{(*)}-2^2-9\\\\=(x+2)^2-4-9=(x+2)^2-13\\\\x^2+4x-9=(x+a)^2+b\\\Downarrow\\(x+2)^2-13=(x+a)^2+b\Rightarrow \boxed{\bold{a=2, b=-13}}$

$\bold{2.}\\\\\underbrace{(x+a)^2}_{(*)}+b=x^2+2ax+a^2+b\\\\x^2+4x-9=x^2+2ax+(a^2+b)\Rightarrow 4=2a\ \text{and}\ -9=a^2+b\\\\4=2a\qquad\text{divide both sides by 2}\\\\\dfrac{4}{2}=\dfrac{2a}{2}\\\\2=a\to \boxed{\bold{a=2}}\\\\\text{Substitute to the second equation:}\\\\-9=2^2+b\\\\-9=4+b\qquad\text{subtract 4 from both sides}\\\\-9-4=4-4+b\\\\-13=b\to \boxed{\bold{b=-13}}$

$\bold{3.}\\\\(x+a)^2+b-\text{it's a vertex form of an equation of a parabola}\\\\\text{Let}\ y=a(x-h)^2+k\ -\ \text{the euation of a parabola}\ y=ax^2+bx+c.\\\\\text{Then}\ h=\dfrac{-b}{2a},\ k=\dfrac{-(b^2-4ac)}{4a}\\\\x^2+4x-9\to a=1,\ b=4,\ c=-9\\\\h=\dfrac{-4}{2(1)}=\dfrac{-4}{2}=-2\\\\k=\dfrac{-(4^2-4(1)(-9))}{4(1)}=\dfrac{-(16+36)}{4}=\dfrac{-42}{4}=-13\\\\\text{Therefore}\\\\x^2+4x-9=(x+a)^2+b\\\\(x-(-2))^2+(-13)=(x+a)^2+b\\\\(x+2)^2-13=(x+a)^2+b\to\boxed{\bold{a=2,\ b=-13}}$