Question

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

2) Solve. x2 – 5x – 6 = 0

3) What value should be added to the expression to create a perfect square? x2 – 20x

4) Solve. x2 + 8x – 8 = 0

5) Solve: 2x2 + 12x = 0

6) Solve each problem by using the quadratic formula. Write solutions in simplest radical form. 2x2 – 2x – 1 = 0

7) Calculate the discriminant. x2 – x + 2 = 0

8) Calculate the discriminant and use it to determine how many real-number roots the equation has. 3x2 – 6x + 1 = 0

9) Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis. y = 2x2 + x – 3


10) Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis. y = x2 – 12x + 12

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=(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 * 1 * (-8)=64+32=96 \\ √(\Delta)=√(96)=√(16 *6)=4√(6) \\ \\ x=(-b \pm √(\Delta))/(2a)=(-8 \pm 4√(6))/(2 * 1)=(2(-4 \pm 2√(6)))/(2)=-4 \pm 2√(6) \\ \boxed{x=-4-2√(6) \hbox{ or } x=-4+2√(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 * 2 * (-1)=4+8=12 \\ √(\Delta)=√(12)=√(4 * 3)=2√(3) \\ \\ x=(-b \pm √(\Delta))/(2a)=(-(-2) \pm 2√(3))/(2 * 2)=(2 \pm 2√(3))/(2 * 2)=(2(1 \pm √(3)))/(2 * 2)=(1 \pm √(3))/(2) \\ \boxed{x=(1-√(3))/(2) \hbox{ or } x=(1+√(3))/(2)}`

7)

`x^2-x+2=0 \\ \\ a=1 \\ b=-1 \\ c=2 \\ \Delta=b^2-4ac=(-1)^2-4 * 1 * 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 * 3 * 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 * 2 * (-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 * 1 * 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}}`