FairyTailsLucy
Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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

Sagot :

naǫ
1)
[tex]x^2+8x+16=0 \\ (x+4)^2=0 \\ x+4=0 \\ \boxed{x=-4}[/tex]

2)
[tex]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}[/tex]

3)
[tex]\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}}[/tex]

4)
[tex]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}}[/tex]

5)
[tex]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}[/tex]

6)
[tex]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}}[/tex]

7)
[tex]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}[/tex]

8)
[tex]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}}[/tex]

9)
[tex]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}}[/tex]

10)
[tex]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}}[/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.