Determine a quadratic equation with the given roots
1. 2 and -1
2. -2 and -6
3. 3 and 4
Answers
Answer:
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Answer :
- The quadratic equation formed by the roots 2 and (-1) is x² - x - 2 = 0
- The quadratic equation formed by the roots (-2) and (-6) is x² + 8x + 12 = 0
- The quadratic equation formed by the roots 3 and 4 is x² - 7x + 12 = 0
Explanation :
Given :
- Roots of the first quadratic equation = 2 and (-1).
- Roots of the second quadratic equation = (-2) and (-6)
- Roots of the second quadratic equation formed = 3 and 4
To find :
- Quadratic equations whose roots are given.
Knowledge required :
Formula for forming a quadratic equation :
⠀⠀⠀⠀⠀⠀⠀⠀⠀x² - (α + β)x + αβ = 0
Where :
- α and β are the two roots of the equation
Solution :
Quadratic equation with roots of 2 and (-1) :
By using the formula for forming a quadratic equation and substituting the values in it, we get :
==> x² - (α + β)x + αβ = 0
==> x² - (2 + (-1))x + 2 × (-1) = 0
==> x² -(2 - 1)x + (-2) = 0
==> x² - x - 2 = 0
∴ Quadratic Equation = x² - x - 2 = 0
Quadratic equation with roots of (-2) and (-6) :
By using the formula for forming a quadratic equation and substituting the values in it, we get :
==> x² - (α + β)x + αβ = 0
==> x² - [(-2) + (-6)]x + (-2) × (-6) = 0
==> x² - (-2 - 6)x + 12 = 0
==> x² - (-8)x + 12 = 0
==> x² + 8x + 12 = 0
∴ Quadratic Equation = x² + 8x + 12 = 0
Quadratic equation with roots of 3 and 4 :
By using the formula for forming a quadratic equation and substituting the values in it, we get :
==> x² - (α + β)x + αβ = 0
==> x² - (3 + 4)x + 3 × 4 = 0
==> x² - 7x + 12 = 0
∴ Quadratic Equation = x² - 7x + 12 = 0
Therefore,
- The quadratic equation formed by the roots 2 and (-1) is x² - x - 2 = 0
- The quadratic equation formed by the roots (-2) and (-6) is x² + 8x + 12 = 0
- The quadratic equation formed by the roots 3 and 4 is x² - 7x + 12 = 0