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Find the zeroes of the polynomial $\displaystyle p(x) = 3x^2 + x - 10$ and verify the relationship between zeroes and its coefficients
Find the zeroes of the polynomial $\displaystyle p(x) = 3x^2 + x - 10$ and verify the relationship between zeroes and its coefficients
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We are given the quadratic polynomial:
$ \displaystyle p(x) = 3x^2 + x - 10 $
### Step 1: Find the Zeroes Using the Quadratic Formula
The roots of a quadratic equation $ ax^2 + bx + c = 0 $ are given by:
$ \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
For $ p(x) = 3x^2 + x - 10 $, we have:
- $ a = 3 $, $ b = 1 $, $ c = -10 $
Substituting these values:
$ \displaystyle x = \frac{-1 \pm \sqrt{(1)^2 - 4(3)(-10)}}{2(3)} $
$ \displaystyle x = \frac{-1 \pm \sqrt{1 + 120}}{6} $
$ \displaystyle x = \frac{-1 \pm \sqrt{121}}{6} $
$ \displaystyle x = \frac{-1 \pm 11}{6} $
Solving for both values:
1. $ \displaystyle x = \frac{-1 + 11}{6} = \frac{10}{6} = \frac{5}{3} $
2. $ \displaystyle x = \frac{-1 - 11}{6} = \frac{-12}{6} = -2 $
Thus, the zeroes of the polynomial are:
$ \displaystyle \frac{5}{3} \quad \text{and} \quad -2 $
### Step 2: Verify the Relationship Between Zeroes and Coefficients
Using Vieta’s formulas:
- Sum of zeroes: $ \displaystyle \alpha + \beta = \frac{-b}{a} $
- Product of zeroes: $ \displaystyle \alpha \beta = \frac{c}{a} $
Substituting the values:
- $ \displaystyle \frac{5}{3} + (-2) = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3} $
- $ \displaystyle \frac{-b}{a} = \frac{-1}{3} $
Since both are equal, the sum is verified.
For the product:
- $ \displaystyle \left( \frac{5}{3} \right) \times (-2) = \frac{-10}{3} $
- $ \displaystyle \frac{c}{a} = \frac{-10}{3} $
Since both are equal, the product is verified.
### Final Answer:
The zeroes of $ p(x) = 3x^2 + x - 10 $ are $ \displaystyle \frac{5}{3} $ and $ -2 $, and the relationships between zeroes and coefficients are verified.
$ \displaystyle p(x) = 3x^2 + x - 10 $
### Step 1: Find the Zeroes Using the Quadratic Formula
The roots of a quadratic equation $ ax^2 + bx + c = 0 $ are given by:
$ \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
For $ p(x) = 3x^2 + x - 10 $, we have:
- $ a = 3 $, $ b = 1 $, $ c = -10 $
Substituting these values:
$ \displaystyle x = \frac{-1 \pm \sqrt{(1)^2 - 4(3)(-10)}}{2(3)} $
$ \displaystyle x = \frac{-1 \pm \sqrt{1 + 120}}{6} $
$ \displaystyle x = \frac{-1 \pm \sqrt{121}}{6} $
$ \displaystyle x = \frac{-1 \pm 11}{6} $
Solving for both values:
1. $ \displaystyle x = \frac{-1 + 11}{6} = \frac{10}{6} = \frac{5}{3} $
2. $ \displaystyle x = \frac{-1 - 11}{6} = \frac{-12}{6} = -2 $
Thus, the zeroes of the polynomial are:
$ \displaystyle \frac{5}{3} \quad \text{and} \quad -2 $
### Step 2: Verify the Relationship Between Zeroes and Coefficients
Using Vieta’s formulas:
- Sum of zeroes: $ \displaystyle \alpha + \beta = \frac{-b}{a} $
- Product of zeroes: $ \displaystyle \alpha \beta = \frac{c}{a} $
Substituting the values:
- $ \displaystyle \frac{5}{3} + (-2) = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3} $
- $ \displaystyle \frac{-b}{a} = \frac{-1}{3} $
Since both are equal, the sum is verified.
For the product:
- $ \displaystyle \left( \frac{5}{3} \right) \times (-2) = \frac{-10}{3} $
- $ \displaystyle \frac{c}{a} = \frac{-10}{3} $
Since both are equal, the product is verified.
### Final Answer:
The zeroes of $ p(x) = 3x^2 + x - 10 $ are $ \displaystyle \frac{5}{3} $ and $ -2 $, and the relationships between zeroes and coefficients are verified.
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