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# Solutions Using the Discriminant

## Identify and use the discriminant

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Practice Solutions Using the Discriminant
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Telescope
Teacher Contributed

## Real World Applications – Algebra I

### Topic

Designing a Mirror for a Telescope in Space

### Student Exploration

Astronomers use telescopes to get a good view to see what’s out there in space. For this activity, we’re going to start to look at the importance of designing a mirror for one of these telescopes and represent its design using quadratics.

In this particular activity, we’re given that the circular area of this telescope is 14.12 square meters. This means that the radius is 2.12 meters, and the diameter is 4.24 meters.

We also know that there is a particular equation that represents the focal length of the parabola of the mirror. This equation is , where represents the focal length and “” represents value of the coefficient of our quadratic equation. If we want our focal length to be 8 meters, we can use this to determine our “” value.

Now we want to multiply both sides by “” to get rid of the denominator of the fraction. Then divide both sides by 32 to isolate the variable.

Now that we have our intercept at (2.12, 0) that represents the radius of the telescope, we are able to input this information into the vertex form of the quadratic and then be able to sketch the graph of our mirror.

The vertex form of a quadratic is , where represents the vertex. In this case, (2.12, 0) is our vertex because the radius of our mirror is 2.12 meters. So, we now have . Let’s simplify the right side of this equation.

Multiply these two binomials first! This is one of those “special” binomials you have to multiply.

Combine like terms in the parentheses.

Now multiply all of the terms on the inside by the term on the outside of the parentheses, and make sure your final equation is in standard form.

Now we have our quadratic equation in standard form, meaning that on the right side of the equation, we look at the higher order of the terms. In this case, is the highest ordered term, and should be first in the list of terms on the right side of the equation.

Also, when looking at this equation, we see that the intercept, (0, 0.14045), represents the vertical length of the mirror.

If we were to graph this quadratic equation, we will have the following:

The axis represents the length of the mirror in meters, and the axis represents the vertical height of the mirror in meters as well.

Now, just by looking at this graph, it doesn’t have the shape of a symmetrical mirror for an astronomer. Why not?

We know that the radius of the mirror is 2.12 meters, and the diameter is 4.24 meters. Let’s change our graph so that our graph only reaches a maximum width, or horizontal distance, of 4.24 meters.

Let’s dig a little bit deeper into the Algebra of this quadratic equation and analyze what the discriminant means.

For the equation, the discriminant is the part under the square root sign when finding the intercept by using the quadratic formula.

When we use the quadratic formula to find the intercepts, we have:

Let’s look at the discriminant, which is the part under the square root. We see that so far, the discriminant is “” which is also zero. What does this mean?

After reviewing this concept, we know that if the discriminant is zero, then there is only one intercept. In this case, our one intercept is the vertex, which is also the radius of this telescope mirror.