Question

1) Solve by using the perfect squares method. x2 + 8x + 16 = 0

Answer 1

1)
$x^2+8x+16=0 \\ (x+4)^2=0 \\ x+4=0 \\ \boxed{x=-4}$

2)
$x^2-5x-6=0 \\ x^2-6x+x-6=0 \\ x(x-6)+1(x-6)=0 \\ (x+1)(x-6)=0 \\ x+1=0 \ \lor \ x-6=0 \\ x=-1 \ \lor \ x=6 \\ \boxed{x=-1 \hbox{ or } x=6}$

3)
$\hbox{a perfect square:} \\ (x-a)^2=x^2-2xa+a^2 \\ \\ 2xa=20x \\ a=\frac{20x}{2x} \\ a=10 \\ \\ a^2=10^2=100 \\ \\ \hbox{the expression:} \\ x^2-20x+100 \\ \\ \boxed{\hbox{100 should be added to the expression}}$

4)
$x^2+8x-8=0 \\ \\ a=1 \\ b=8 \\ c=-8 \\ \Delta=b^2-4ac=8^2-4 \times 1 \times (-8)=64+32=96 \\ \sqrt{\Delta}=\sqrt{96}=\sqrt{16 \times6}=4\sqrt{6} \\ \\ x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-8 \pm 4\sqrt{6}}{2 \times 1}=\frac{2(-4 \pm 2\sqrt{6})}{2}=-4 \pm 2\sqrt{6} \\ \boxed{x=-4-2\sqrt{6} \hbox{ or } x=-4+2\sqrt{6}}$

5)
$2x^2+12x=0 \\ 2x(x+6)=0 \\ 2x=0 \ \lor \ x+6=0 \\ x=0 \ \lor \ x=-6 \\ \boxed{x=-6 \hbox{ or } x=0}$

6)
$2x^2-2x-1=0 \\ \\ a=2 \\ b=-2 \\ c=-1 \\ \Delta=b^2-4ac=(-2)^2-4 \times 2 \times (-1)=4+8=12 \\ \sqrt{\Delta}=\sqrt{12}=\sqrt{4 \times 3}=2\sqrt{3} \\ \\ x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-2) \pm 2\sqrt{3}}{2 \times 2}=\frac{2 \pm 2\sqrt{3}}{2 \times 2}=\frac{2(1 \pm \sqrt{3})}{2 \times 2}=\frac{1 \pm \sqrt{3}}{2} \\ \boxed{x=\frac{1-\sqrt{3}}{2} \hbox{ or } x=\frac{1+\sqrt{3}}{2}}$

7)
$x^2-x+2=0 \\ \\ a=1 \\ b=-1 \\ c=2 \\ \Delta=b^2-4ac=(-1)^2-4 \times 1 \times 2=1-8=-7 \\ \\ \boxed{\hbox{the discriminant } \Delta=-7}$

8)
$3x^2-6x+1=0 \\ \\ a=3 \\ b=-6 \\ c=1 \\ \Delta=b^2-4ac=(-6)^2-4 \times 3 \times 1=36-12=24 \\ \\ \boxed{\hbox{the discriminant } \Delta=24} \\ \\ \hbox{if } \Delta\ \textless \ 0 \hbox{ then there are no real roots} \\ \hbox{if } \Delta=0 \hbox{ then there's one real root} \\ \hbox{if } \Delta\ \textgreater \ 0 \hbox{ then there are two real roots} \\ \\ \Delta=24\ \textgreater \ 0 \\ \boxed{\hbox{the equation has two real roots}}$

9)
$y=2x^2+x-3 \\ \\ a=2 \\ b=1 \\ c=-3 \\ \Delta=b^2-4ac=1^2-4 \times 2 \times (-3)=1+24=25 \\ \\ \hbox{the function has two zeros} \\ \boxed{\hbox{the parabola has 2 points in common with the x-axis}} \\ \\ a\ \textgreater \ 0 \hbox{ so the parabola ope} \hbox{ns upwards} \\ \boxed{\hbox{the vertex lies below the x-axis}}$

10)
$y=x^2-12x+12 \\ \\ a=1 \\ b=-12 \\ c=12 \\ \Delta=b^2-4ac=(-12)^2-4 \times 1 \times 12=144-48=96 \\ \\ \hbox{the function has two zeros} \\ \boxed{\hbox{the parabola has 2 points in common with the x-axis}} \\ \\ a\ \textgreater \ 0 \hbox{ so the parabola ope} \hbox{ns upwards} \\ \boxed{\hbox{the vertex lies below the x-axis}}$