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3 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

Mathematics

2 answers:

vodomira [7]3 years ago

7 0

Finding the discriminate of a quadratic formula determines the number and type of answers.

The formula is b^2 -4(ac)

a = 6. b = -7 and c = -4

-7^2 -4(6*-4) = 145

The answer is a positive number so this means there are 2 real solutions.

stepladder [879]3 years ago

4 0

Answer:

two real solutions

Step-by-step explanation:

We can use the discriminant to determine the type and number of solutions

b^2 -4ac where ax^2 +bx +c

If b^2 -4ac > 0 we have 2 real solutions

b^2 -4ac = 0 there is 1 real solution

b^2 -4ac < 0 there are 2 complex solutions

6x^2 -7x -4 = 0

a=6 b =-7 c = -4

b^2 -4ac

(-7)^2 -4(6)(-4)

49 + 96

145>0

There are real solutions

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Differential Equations Problem

fgiga [73]

Answer:

$y=\frac{t^2e^{2t}}{3}+ce^{2t}$

Step-by-step explanation:

We have given differential equation $\frac{dy}{dt}-2y=t^2e^{2t}$

We know that linear differential equation is given by $\frac{dy}{dt}+Py=Q$

On comparing with standard equation P = -2 and Q= $t^2e^{2t}$

Now integrating factor $IF=e^{-Pdt}$

$IF=e^{-2dt}=e^{-2t}$

Now solution of differential equation is given by

$y\times IF=\int\ IF\times Q\ dt$

$y\times e^{-2t}=\int\ e^{-2t}\times t^2e^{2t}\ dt$

$y\times e^{-2t}=\frac{t^2}{3}+c$

$y=\frac{t^2e^{2t}}{3}+ce^{2t}$

7 0

3 years ago

Find the root of (M^3)+(M^2)-2=0. Show steps.

MrRissso [65]

Answer:

x = 1, -1 + i, and -1 - i are the 3 roots

Step-by-step explanation:

The Rational Root Theorem will help us find any real roots in this 3rd degree polynomial. The c value in that polynomia is -2 and the d value is 1 (the leading coefficient).

If c is -2, then the factors of -2 are:

1, -1, 2, -2

If d is 1, then the factors of 1 are:

1, -1

And c/d gives us all the possible roots:

1/1, -1/1, 2/1, -2/1 which of course simplify to

1, -1 2, -2

We will now use synthetic division to find the first real root. I will use x = 1 first.

For synthetic division:

1| 1 1 0 -2

is how it is set up. Bring down the first number, 1, multiply it by 1 and put that product up under the second "1":

1| 1 1 0 -2

Now add to get

1| 1 1 0 -2

1 2

Multiply 2 by 1 and put that product up under the 0 and add:

1| 1 1 0 -2

1 2

1 2 2

Again multiply the 2 by 1 and put that product up undeer the -2 and add:

1| 1 1 0 -2

1 2 2

1 2 2 0

That remainder of 0 tells us that x = 1 is a root of that polynomial and the depressed polynomial, made up from the bold numbers, is one degree less than what it started with:

$x^2+2x+2=0$