What does x equal in x+4=2x-3

Consider the quadratic equation ax^2+bx+5=0,where a, b and c are rational numbers and the quadratic has two distinct zeros.

FrozenT [24]

You have shared the situation (problem), except for the directions: What are you supposed to do here? I can only make a educated guesses. See below:

Note that if   <span>ax^2+bx+5=0 then it appears that c = 5 (a rational number).

Note that for simplicity's sake, we need to assume that the "two distinct zeros" are real numbers, not imaginary or complex numbers. If this is the case, then the discriminant, b^2 - 4(a)(c), must be positive. Since c = 5,

b^2 - 4(a)(5) > 0, or b^2 - 20a > 0.

Note that if the quadratic has two distinct zeros, which we'll call "d" and "e," then

(x-d) and (x-e) are factors of ax^2 + bx + 5 = 0, and that because of this fact,

- b plus sqrt( b^2 - 20a )
d =  ------------------------------------
2a

and

</span> - b minus sqrt( b^2 - 20a )
e =  ------------------------------------
2a

Some (or perhaps all) of these facts may help us find the values of "a" and "b." Before going into that, however, I'm asking you to share the rest of the problem statement. What, specificallyi, were you asked to do here?

7 0

3 years ago

The sum of a number, 1/6 of that number, 2 1/2 of that number, and 7 is 12 1/2. Find the number.

Tema [17]

Answer:

2 $\frac{1}{16}$

Step-by-step explanation:

The question state that; the sum of a number, 1/6 of that number, 2 1/2 of that number, and 7 is 12 1/2. Find the number.

First, we need to interpret and represent this statement mathematically

Let n be the number.

The sum means 'addition'

1/6 of that number is 1/6 of n = 1/6 × n = $\frac{n}{6}$

2 1/2 of that number is 2 1/2 of n = 2 1/2 × n = 5/2 × n = $\frac{5n}{2}$

So the question says that, when we add; $\frac{n}{6}$ , $\frac{5n}{2}$ and 7 all together, the value is 12 1/2

That is;

$\frac{n}{6}$ + $\frac{5n}{2}$ + 7 = 12 1/2

$\frac{n}{6}$ + $\frac{5n}{2}$ + 7 = $\frac{25}{2}$ ( changing 12 1/2 to improper fraction will give $\frac{25}{2}$ )

So we can now go ahead and solve for n

$\frac{n}{6}$ + $\frac{5n}{2}$ + 7 = $\frac{25}{2}$

subtract 7 from both-side of the equation

$\frac{n}{6}$ + $\frac{5n}{2}$ = $\frac{25}{2}$ - 7

$\frac{n + 15n}{6}$ = $\frac{25 -14}{2}$

$\frac{16n}{6}$ = $\frac{11}{2}$

$\frac{16n}{6}$ can be reduced to give us $\frac{8n}{3}$

$\frac{8n}{3}$ = $\frac{11}{2}$

Cross multiply

8n × 2 = 11× 3

16n = 33

Divide both-side of the equation by 16

$\frac{16n}{16}$ = $\frac{33}{16}$

(On the left-hand side of the equation, 16 will cancel-out 16 leaving us with just, while on the right-hand side of the equation 33 will be divided by 16)

n = $\frac{33}{16}$ = $2\frac{1}{16}$