What kind of solutions does 7R2 -14R + 10 = 0 have?

Jul 5, 2015

$7 {R}^{2} - 14 R + 10$ has discriminant $\Delta = - 84 < 0$.

So $7 {R}^{2} - 14 R + 10 = 0$ has no real solutions.

It has two distinct complex solutions.

Explanation:

$7 {R}^{2} - 14 R + 10$ is of the form $a {R}^{2} + b R + c$ with $a = 7$, $b = - 14$ and $c = 10$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 14\right)}^{2} - \left(4 \times 7 \times 10\right) = 196 - 280 = - 84$

Since $\Delta < 0$ the equation $7 {R}^{2} - 14 R + 10 = 0$ has no real roots. It has a pair of complex roots that are complex conjugates of one another.

The possible cases are:
$\Delta > 0$ The quadratic equation has two distinct real roots. If $\Delta$ is a perfect square (and the coefficients of the quadratic are rational), then those roots are also rational.

$\Delta = 0$ The quadratic equation has one repeated real root.

$\Delta < 0$ The quadratic equation has no real roots. It has a pair of distinct complex roots that are complex conjugates of one another.