If α and β are roots of x^2-4x+3=0, which equation has 3α and 3β as roots?
Answer and explanation
Correct answer: x^2-12x+27=0
For the original equation, Vieta’s relations give α+β=4 and αβ=3. If the new roots are 3α and 3β, their sum is 3α+3β=3(α+β)=12, while their product is (3α)(3β)=9αβ=27. A monic quadratic with roots r and s is x^2-(r+s)x+rs=0. Substituting the new sum and product gives x^2-12x+27=0, so option A is correct. Option B keeps the old sum, option C uses the old product multiplied by only 3 instead of 9, and option D has the wrong sign for the x-term. The result can also be checked from the original roots 1 and 3, which become 3 and 9.
Frequently asked questions
What is the correct answer to this question?
x^2-12x+27=0
Why is this the correct answer?
For the original equation, Vieta’s relations give α+β=4 and αβ=3. If the new roots are 3α and 3β, their sum is 3α+3β=3(α+β)=12, while their product is (3α)(3β)=9αβ=27. A monic quadratic with roots r and s is x^2-(r+s)x+rs=0. Substituting the new sum and product gives x^2-12x+27=0, so option A is correct. Option B keeps the old sum, option C uses the old product multiplied by only 3 instead of 9, and option D has the wrong sign for the x-term. The result can also be checked from the original roots 1 and 3, which become 3 and 9.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
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