Question

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

Answer 1

`\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

METHOD 1:

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`

METHOD 2:

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`