Best Answer:
Let's assume there is at least one solution:

x² + 6x + 10 = 0

Remember the discriminant of a quadratic tells you how many solutions there are. The discriminant is that portion of the quadratic formula under the square root symbol.

d = b² - 4ac

Rules:

If d > 0, there are 2 real solutions

If d = 0, there is 1 real solution (double root)

If d < 0, there are no real solutions

Indeed if you plug in your values for a, b, c you get a negative discriminant:

d = b² - 4ac

a = 1

b = 6

c = 10

d = 6² - 4(1)(10)

d = 36 - 40

d = -4

Hence there are no real solutions.

Another way you can show this is to rewrite the equation for the parabola in vertex form:

y = (x - h)² + k

We have:

y = x² + 6x + 10

To complete the square:

a. Take the coefficient in front of x --> 6

b. Halve it --> 3

c. Square it --> 3² = 9

d. Add it and subtract it:

x² + 6x + 9 - 9 + 10

The first part is a perfect square:

(x² + 6x + 9) + (-9 + 10)

(x + 3)(x + 3) + 1

y = (x + 3)² + 1

In vertex form we know that (-3, 1) is the vertex and since it is upward facing (leading coefficient is positive), it will never go below the x-axis.

Or you could mathematically show that this can never equal zero.

As you can see the square can never be negative. And if you add 1 to that, it will always be positive. It can't ever be zero.

You can perhaps see this more clearly, if you try to set it to zero.

(x + 3)² + 1 = 0

(x + 3)² = -1

A square of a real number can never be negative, so there is no real number solution for x.

A third way would be to graph y = x² + 6x + 10 and show that it never crosses the x-axis, so there is no way it could ever be equal to zero.

Can you think of other ways to prove this? Which way do you prefer?

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