MATH SOLVE

6 months ago

Q:
# The difference between the roots of the quadratic equation x2+x+c=0 is 6. Find c.

Accepted Solution

A:

We have that

[tex]x_{1} - x_{2} = 6[/tex]

Let y be the first root, then y - 6 will be the second one. Since both roots make the quadratic equation equal to 0, we have that:

[tex] y^{2} + y + c = (y - 6)^{2} + (y - 6) +c[/tex]

Solve for y:

[tex]y^{2} +y +c = y^{2} - 12y + 36 + y - 6 +c \\ y + 12y - y = 30 \\ y = \frac{5}{2} [/tex]

Plug in y = 5/2 into original quadratic equation and solve for c:

[tex]( \frac{5}{2}) ^{2} + \frac{5}{2} +c = 0 \\ c= - \frac{5}{2} - \frac{25}{4} \\ c = - \frac{35}{4} [/tex]

So, the answer is c = -35/4

[tex]x_{1} - x_{2} = 6[/tex]

Let y be the first root, then y - 6 will be the second one. Since both roots make the quadratic equation equal to 0, we have that:

[tex] y^{2} + y + c = (y - 6)^{2} + (y - 6) +c[/tex]

Solve for y:

[tex]y^{2} +y +c = y^{2} - 12y + 36 + y - 6 +c \\ y + 12y - y = 30 \\ y = \frac{5}{2} [/tex]

Plug in y = 5/2 into original quadratic equation and solve for c:

[tex]( \frac{5}{2}) ^{2} + \frac{5}{2} +c = 0 \\ c= - \frac{5}{2} - \frac{25}{4} \\ c = - \frac{35}{4} [/tex]

So, the answer is c = -35/4