Use the discriminant to determine the number of real solutions of x2+10x=−21

.(1 point)

1 answer

To use the discriminant to determine the number of real solutions of the equation \(x^2 + 10x = -21\), we first need to rewrite the equation in standard form:

\[ x^2 + 10x + 21 = 0 \]

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). Here, we have:

  • \(a = 1\)
  • \(b = 10\)
  • \(c = 21\)

The discriminant \(\Delta\) is given by the formula:

\[ \Delta = b^2 - 4ac \]

Now, we can substitute the values of \(a\), \(b\), and \(c\) into the discriminant formula:

\[ \Delta = 10^2 - 4(1)(21) \] \[ \Delta = 100 - 84 \] \[ \Delta = 16 \]

Now we analyze the value of the discriminant:

  • If \(\Delta > 0\), there are 2 distinct real solutions.
  • If \(\Delta = 0\), there is 1 real solution (a repeated solution).
  • If \(\Delta < 0\), there are no real solutions.

Since \(\Delta = 16\) which is greater than 0, we can conclude that the equation \(x^2 + 10x + 21 = 0\) has 2 distinct real solutions.