Question

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

Answer 1

We have that
$x_{1} - x_{2} = 6$
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:
$y^{2} + y + c = (y - 6)^{2} + (y - 6) +c$
Solve for y:
$y^{2} +y +c = y^{2} - 12y + 36 + y - 6 +c \\ y + 12y - y = 30 \\ y = \frac{5}{2}$
Plug in y = 5/2 into original quadratic equation and solve for c:
$( \frac{5}{2}) ^{2} + \frac{5}{2} +c = 0 \\ c= - \frac{5}{2} - \frac{25}{4} \\ c = - \frac{35}{4}$
So, the answer is c = -35/4

Answer 2

To solve this question:

• First, we compare with the standard quadratic equation to find a, b and c.
• Then, we apply Bhaskara to find the roots, as function of c.
• Then, since the difference is 6, we use this to find c.

Doing this, we get that: $c = -\frac{35}{4}$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

x²+x+c=0

Comparing with the standard second order equation, which is:

$ax^2 + bx + c = 0$

We get that:

$a = 1, b = 1$

Finding the roots:

$\Delta = b^{2} - 4ac = 1^2 - 4(1)(c) = 1 - 4c$

$x_{1} = \frac{-1 + \sqrt{1 - 4c}}{2}$

$x_{2} = \frac{-1 - \sqrt{1 - 4c}}{2}$

Difference of 6

Thus:

$x_1 - x_2 = 6$

$\frac{-1 + \sqrt{1 - 4c}}{2} - \frac{-1 - \sqrt{1 - 4c}}{2} = 6$

$-\frac{1}{2} + \frac{\sqrt{1 - 4c}}{2} + \frac{1}{2} + \frac{\sqrt{1 - 4c}}{2} = 6$

$\frac{2\sqrt{1 - 4c}}{2} = 6$

$\sqrt{1 - 4c} = 6$

$(\sqrt{1 - 4c})^2 = 6^2$

$1 - 4c = 36$

$-4c = 35$

$4c = -35$

$c = -\frac{35}{4}$

Thus, $c = -\frac{35}{4}$ is the value.

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