The difference of the two solutions of x^2+19x+c=0 is 15. Find c.

The value of is 34.

It is given that,

Explanation:

Let and are two solutions of the given equation.

          ...(i)

The sum of solutions of is . So,

         ...(ii)

Adding (i) and (ii), we get

Substitute in (ii).

The product of solutions of is . So,

Thus, the value of is 34.

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34

Step-by-step explanation:

For a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a, and the product of the roots is c/a.

If we say the roots are p and q, then:

p + q = -19/1

pq = c/1

The difference of the roots is 15:

p − q = 15

Add to the first equation:

2p = -4

p = -2

Which means the value of q is -17.

So the value of c is (-2)(-17) = 34.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3