Convert to vertex form 3x^2+6x-5=0

Question

asked Mar 18, 2021 22.3k views

1 Answer

Wrek · Answer 1 · 2021-03-23T12:08:04+0000

Answer:

$\large\boxed{3x^2+6x-5=0\Rightarrow3(x+1)^2-8=0}$

Explanation:

The vertex form of an equation of a quadratic equation

ax^2+bx+c=0 :

a(x-h)^2+k=0

(h, k) - vertex

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

$3x^2+6x-5=0\\\\3x^2+2(x)(3)(1)-5=0\\\\3x^2+(3)(2)(x)(1)+(3)(1^2)-(3)(1^2)-5=0\\\\3\underbrace{(x^2+2(x)(1)+1^2)}_((*))-(3)(1)-5=0\\\\3(x+1)^2-8=0$

Use the formula:

$h=(-b)/(2a),\ k=ah^2+bh+c$

Substitute:

$3x^2+6x-5\to a=3,\ b=6,\ c=-5\\\\h=(-6)/((2)(3))=(-6)/(6)=-1\\\\k=3(-1)^2+6(-1)-5=3-6-5=-8$

$a(x-h)^2+k:\\\\3(x-(-1))^2+(-8)=3(x+1)^2-8$