HELP!!! Solve by using the Quadratic Equation: 1. x²+10x+25=0 2. 5x²+2x+10=0 3. 3x²=7x+20
Let \(f(x)=ax^2+bx+c\). Then\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\]
I know that equation, but im still not getting it.
1. \((1)x^2+(10)x+(25)=0\),\[x=\frac{-10\pm\sqrt{10^2-4(1)(25)}}{2(1)}=?\]
where did you get the 4 in that problem?
and the 2 on the bottom ?
\[x=\frac{-b\pm\sqrt{b^2-\color{red}{4}ac}}{\color{red}{2}a}.\]They're part of the equation.
The 4 and two is part of the formula. It is always there when you use this formula. You just have to remember it
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)
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 \(\Large x = -5\)
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
1. -5 2. (-2/10) 3. (-7/6) ??
#1 is correct, #2 and #3 are not
how did you get -2/10 for #2?
i kinda did it how you did number one il draw them out on how i found them, just give me a second
alright
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
In #2, what is the value of a, b, and c?
Number 2 they are a: 5 b: 2 c: 10 Number 3 they are a:3 b: 7 c: 20
#2 you are correct, a = 5, b = 2, and c = 10 #3, that's not correct
would 3 actually be a: 7 b: 3 c: 20
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?
a=3 b=7 c=20 ??
\[\Large 3x^2 - 7x - 20 = 0\] is the same as \[\Large 3x^2 + (-7)x + (-20) = 0\]
Match that up with \[\Large ax^2 + bx + c = 0\]
|dw:1341947383288:dw|
Join our real-time social learning platform and learn together with your friends!