3 years ago

What’s is the product of the two solutions of the quadratic equation ax^2+bx+c=0

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

7 0

The product of the two roots is c/a.

_____

Consider the equation

... a(x -p)(x -q) = 0

which has solutions x=p and x=q.

When multiplied out, it becomes ...

... a(x² -(p+q)x +pq) = ax² -a(p+q)x +apq = 0

When apq = c, the product pq is c/a.

You might be interested in

Solve for x<br><br>4x+6-7+9=18

Alla [95]

here is your ans have a nice day

6 0

3 years ago

Read 2 more answers

How much acetic acid is in a 5 - liter container of acetic acid and water that is marked 80% acetic acid ? How much is water ?

Goryan [66]

I think the question ur asking is what’s the volume of water?? If so then the volume of water is 1L

6 0

1 year ago

Traveling at 65 miles per hour, how many minutes, rounded to the nearest whole number, does it take to drive 125 miles from San

telo118 [61]

It should take 115 minutes

7 0

3 years ago

Read 2 more answers

The water level at Lake Penny changed −1 1/10 in. this year. Next year, it is predicted to change another −1 2/5

frutty [35]

- 1 1/10 + (-1 2/5)

- 1 1/10 + (-1 4/5)

-2 5/10

-2 1/2

7 0

3 years ago

The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional

jolli1 [7]

Answer:

x= $\frac{16}{\sqrt[3]{2}+1 }$

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = $\frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}$

= $\frac{I}{x^2}+\frac{2I}{(16-x)^2}$

now Q' = 0

⇒I ${-\frac{2}{x^3}}+\frac{4}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

[/tex] $\frac{1}{x^3} = \frac{2}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

is the required point

5 0

2 years ago