Question

Completing the Square- Color By Number Thanksgiving Edition

Answer 1

Answer: The roots of question 1) -6 , 2 (Brown)

2) -7, 1 (light brown)

3) 3, -1 (yellow)

4)
`(2\pm √(19) )/(3)` (orange)

5)
`3\pm √(15)` (Red)

6)
`(-5\pm3√(5) )/(2)` (Brown)

Explanation:

Since, First quadratic equation is,
`x^2+4x-12=0`

⇒
`x^2+6x-2x-12=0`

⇒
`x(x+6)-2(x+6)=0`⇒
`(x-2)(x+6)=0`

Therefore, root of
`x^2+4x-12=0` are x = 2, -6

Now,Second quadratic equation is,
`2x^2+12x-14=0`

⇒
`2x^2+14x-2x-14=0`

⇒
`2x(x+7)-2(x+7)=0`⇒
`(2x-2)(x+7)=0`

Therefore, root of
`2x^2+12x-14=0` are x = 1, -7

Third quadratic equation is,
`x^2-2x-3=0`

⇒
`x^2-3x+x-3=0`

⇒
`x(x-3)+1(x-3)=0`⇒
`(x-3)(x+1)=0`

Therefore, root of
`2x^2+12x-14=0` are x = -1, 3

fourth quadratic equation is,
`3x^2-4x-5=0`

By applying quadratic formula, roots of the equation
`3x^2-4x-5=0`are

`x=(2\pm 19)/(3)` (
`x=(-b\pm√(b^2-4ac))/(2a )`)

Similarly, By the quadratic formula,

The roots of
`x^2-6x+4=0` are
`3\pm √(15)`

And, roots of
`3x^2-4x-5=0` are
`(-5\pm3√(5) )/(2)`