The Discriminant Worksheet for 9th Grade | Lesson Planet - Free Printable

It looks like you've uploaded a worksheet titled "Solve the Quadratic", which contains several problems related to quadratic equations. Since I can't view images directly, I’ll help you solve and explain each problem based on the text you provided.

Let’s go through the problems one by one and solve them step-by-step.

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Problem 1: Find the nature of roots of the following quadratic equations

We use the discriminant formula:

> $ D = b^2 - 4ac $

- If $ D > 0 $: Two distinct real roots
- If $ D = 0 $: One real root (repeated)
- If $ D < 0 $: No real roots (complex)

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#### a) $ x^2 + 5x + 6 = 0 $

Here, $ a = 1 $, $ b = 5 $, $ c = 6 $

$$
D = 5^2 - 4(1)(6) = 25 - 24 = 1
$$

Since $ D = 1 > 0 $ → Two distinct real roots

✔ Answer: Two distinct real roots

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#### b) $ x^2 + 4x + 4 = 0 $

$ a = 1 $, $ b = 4 $, $ c = 4 $

$$
D = 4^2 - 4(1)(4) = 16 - 16 = 0
$$

Since $ D = 0 $ → One real root (repeated)

✔ Answer: One real root (repeated)

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#### c) $ x^2 + 3x + 5 = 0 $

$ a = 1 $, $ b = 3 $, $ c = 5 $

$$
D = 3^2 - 4(1)(5) = 9 - 20 = -11
$$

Since $ D = -11 < 0 $ → No real roots (complex)

✔ Answer: No real roots

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Problem 2: Solve the following equations by factoring

#### a) $ 11x + 10 = 0 $

Wait — this is not a quadratic equation. It's linear.

But let's solve it anyway:

$$
11x = -10 \Rightarrow x = -\frac{10}{11}
$$

✔ Answer: $ x = -\frac{10}{11} $

> Note: This might be a typo. Maybe it was meant to be $ x^2 + 11x + 10 = 0 $? But as written, it's linear.

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#### b) $ x^2 - 10x + 25 = 0 $

This is a perfect square trinomial.

$$
x^2 - 10x + 25 = (x - 5)^2 = 0
\Rightarrow x = 5
$$

✔ Answer: $ x = 5 $ (double root)

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Problem 3: Solve the following equations using any method

#### a) $ x^2 + 7x + 10 = 0 $

Factor:
$$
x^2 + 7x + 10 = (x + 2)(x + 5) = 0
\Rightarrow x = -2, -5
$$

✔ Answer: $ x = -2, -5 $

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#### b) $ x^2 - 4x + 4 = 0 $

Perfect square:
$$
(x - 2)^2 = 0 \Rightarrow x = 2
$$

✔ Answer: $ x = 2 $ (repeated root)

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Problem 4: Which equations have real, rational roots that are equal?

From earlier:

- $ x^2 + 4x + 4 = 0 $ → $ D = 0 $, so repeated real rational root → ✔ Yes
- $ x^2 - 4x + 4 = 0 $ → same as above → ✔ Yes

So both (b) and (d) have real, rational, equal roots.

✔ Answer: Equations b and d

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Problem 5: The roots of a quadratic equation can be real...

This seems incomplete. Probably asking: *"The roots of a quadratic equation can be real, irrational, or complex."*

But let’s assume it's a fill-in-the-blank.

✔ Answer: The roots of a quadratic equation can be real, irrational, or complex.

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Problem 6: Can (2+√3, 2−√3) be the roots of a quadratic equation?

Yes!

Because they are conjugate irrational roots, and if a quadratic has rational coefficients, irrational roots come in conjugate pairs.

Let’s form the equation:

Sum of roots: $ (2+\sqrt{3}) + (2-\sqrt{3}) = 4 $

Product: $ (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 $

So the quadratic is:

$$
x^2 - (\text{sum})x + (\text{product}) = x^2 - 4x + 1 = 0
$$

✔ Yes, these can be roots.

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Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1a | Two distinct real roots |
| 1b | One real root (repeated) |
| 1c | No real roots |
| 2a | $ x = -\frac{10}{11} $ |
| 2b | $ x = 5 $ |
| 3a | $ x = -2, -5 $ |
| 3b | $ x = 2 $ |
| 4 | Equations b and d |
| 5 | Real, irrational, or complex |
| 6 | Yes |

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