Question

HELP!!! Solve by using the Quadratic Equation:
1. x²+10x+25=0

2. 5x²+2x+10=0

3. 3x²=7x+20

Answer 1

Let \(f(x)=ax^2+bx+c\). Then\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\]

Answer 2

I know that equation, but im still not getting it.

Answer 3

1. \((1)x^2+(10)x+(25)=0\),\[x=\frac{-10\pm\sqrt{10^2-4(1)(25)}}{2(1)}=?\]

Answer 4

where did you get the 4 in that problem?

Answer 5

and the 2 on the bottom ?

Answer 6

\[x=\frac{-b\pm\sqrt{b^2-\color{red}{4}ac}}{\color{red}{2}a}.\]They're part of the equation.

Answer 7

The 4 and two is part of the formula. It is always there when you use this formula. You just have to remember it

Answer 8

so they would look like this:

1. x= -10+√10²-4(1)(25)
-----------------
2(1)

2. x= -2+√2²-4(5)(10)
---------------
2(5)

3. x= -7+√7²-4(3)(20)
----------------
2(3)

Answer 9

Let's solve the first equation \(\Large x^2+10x+25 = 0\) using the quadratic formula


Note: in the case of \(\Large x^2+10x+25 = 0\), a = 1, b = 10 and c = 25


\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(10)\pm\sqrt{(10)^2-4(1)(25)}}{2(1)}\]


\[\Large x = \frac{-10\pm\sqrt{100-(100)}}{2}\]


\[\Large x = \frac{-10\pm\sqrt{0}}{2}\]


\[\Large x = \frac{-10+\sqrt{0}}{2} \ \text{or} \ x = \frac{-10-\sqrt{0}}{2}\]


\[\Large x = \frac{-10+0}{2} \ \text{or} \ x = \frac{-10-0}{2}\]


\[\Large x = \frac{-10}{2} \ \text{or} \ x = \frac{-10}{2}\]


\[\Large x = -5 \ \text{or} \ x = -5\]


\[\Large x = -5\]


So the only solution to \(\Large x^2+10x+25 = 0\) is \(\Large x = -5\)

Answer 10

I'm going to derive it for you so that you can see. :P

Given \(f(x)=ax^2+bx+c\), let \(f(x)=0\). Then\[ax^2+bx+c=0,\]\[x^2+\frac{b}{a}x=-\frac{c}{a},\]\[x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a},\]\[\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a},\]\[x+\frac{b}{2a}=\pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}},\]\[x=-\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}},\]\[x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}},\]\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\]lol

Answer 11

1. -5
2. (-2/10)
3. (-7/6)

??

Answer 12

#1 is correct, #2 and #3 are not

Answer 13

how did you get -2/10 for #2?

Answer 14

i kinda did it how you did number one il draw them out on how i found them, just give me a second

Answer 15

alright

Answer 16

2. -(2)+√4-(4)
-----------
10
which you get -2
----
10

3. -(7)+√49-(49)
----------------
6

which you get -7
-------
6


if this is wrong then can you explain it to me in another way, because i dont get this, i got number one by putting it in my calculator, but the other 2 wouldnt come out as a number

Answer 17

In #2, what is the value of a, b, and c?

Answer 18

Number 2 they are
a: 5
b: 2
c: 10


Number 3 they are
a:3
b: 7
c: 20

Answer 19

#2 you are correct, a = 5, b = 2, and c = 10

#3, that's not correct

Answer 20

would 3 actually be
a: 7
b: 3
c: 20

Answer 21

You have to get everything to one side


\[\Large 3x^2 = 7x + 20\]


\[\Large 3x^2 - 7x = 20\]


\[\Large 3x^2 - 7x - 20 = 0\]


Now what are a, b, and c?

Answer 22

a=3
b=7
c=20

??

Answer 23

\[\Large 3x^2 - 7x - 20 = 0\]

is the same as

\[\Large 3x^2 + (-7)x + (-20) = 0\]

Answer 24

Match that up with

\[\Large ax^2 + bx + c = 0\]