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3 years ago

A quadratic equation is shown below:

1 answer:

6 0

Part A:
3x^2-15x+20=0
ax^2+bx+c=0; a=3, b=-15, c=20
Radicand: R=b^2-4ac
R=(-15)^2-4(3)(20)
R=225-240
R=-15<0, then the equation has two imaginary solutions (no real)

Part B:
3x^2+5x-8=0
a=3, b=5, c=-8
R=(5)^2-4(3)(-8)
R=25+96
R=121>0, then the equation has two different real solutions:
x=[-b+-sqrt(R)] / (2a)
x=[-5+-sqrt(121)] / [2(3)]
x=(-5+-11) / 6

x1=(-5-11)/6=(-16)/6→x1=-8/3

x2=(-5+11)/6=6/6→x2=1

Solutions: x=-8/3 and x=1

I shose this method because I can get the solutions directly, and I don't have to guess the possible solutions

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Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago

PLEASE HELP ME! I really need help y'all...

ikadub [295]

just a quick clarification, tis usually -4.9 and that's a rounded number to reflect earth's gravity on an object in motion, but -5 is close enough :)

$\bf ~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf h(x)=-5(\stackrel{\mathbb{F~O~I~L}}{x^2-8x+16})+180\implies h(x)=-5x^2+40x-80+180 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x)=-5x^2+\stackrel{\stackrel{v_o}{\downarrow }}{40} x+\stackrel{\stackrel{h_o}{\downarrow }}{\boxed{100}}~\hfill$

8 0

4 years ago

Read 2 more answers

yarga [219]

Answer:45

Step-by-step explanation:

6 0

3 years ago

Which graph represents the function y= 2/3x-2?<br>

Volgvan

I believe it’s the 3rd one

8 0

3 years ago

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Jeff needs to buy food and medication for six cats and one dog. The cost of food and medication for a cat is $5 less than two-th

DerKrebs [107]

Let x = amount for dog
let 2/3x-5 = amount for cat

x + 2/3x-5 = 195 combine x's
1 2/3x or 5/3x
5/3x-5 =195 add 5 to both sides
5/3x = 200 multiply both sides by 3
5x = 600 divided both sides by 5
x = 120

the cost of the dog is $120.00

4 0

3 years ago