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) $\frac{2\pm \sqrt{19} }{3}$ (orange)

5)$3\pm \sqrt{15}$ (Red)

6)$\frac{-5\pm3\sqrt{5} }{2}$  (Brown)

Step-by-step 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

$x=\frac{2\pm 19}{3}$     ($x=\frac{-b\pm\sqrt{b^2-4ac}}{2a }$)

Similarly, By the quadratic formula,

The roots of $x^2-6x+4=0$ are $3\pm \sqrt{15}$

And, roots of $3x^2-4x-5=0$ are $\frac{-5\pm3\sqrt{5} }{2}$