1 Quadratic Equations
Which of the following quadratic equation has the sum of their roots 4 and the sum of the cubes of their roots as 28
1) x2−4x+3=0
2) x2−4x−5=0
3) x2−3x+4=0
4) x2+4x+3=0
Let α and β be the roots of the equation. Hence α+β=4 and α3+β3=28 Now α3+β3 can be written as
=(α+β)3−3αβ(α+β) Hence 28=64−12αβ 12αβ=36 αβ=3 Therefore, x2−(α+β)x+αβ=0 x2−(4x)+3=0
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