Which of the following quadratic equation has the sum of their roots $4$ and the sum of the cubes of their roots as $28$ ?

x^{2} - 4 x + 3 = 0

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x^{2} - 4 x - 5 = 0

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x^{2} - 3 x + 4 = 0

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x^{2} + 4 x + 3 = 0

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Solution

The correct option is A $x^{2} - 4 x + 3 = 0$
Let $α$ and $β$ be the roots of the equation.
Hence
$α + β = 4$
and $α^{3} + β^{3} = 28$
Now $α^{3} + β^{3}$ can be written as
$= (α + β)^{3} - 3 α β (α + β)$
Hence
$28 = 64 - 12 α β$
$12 α β = 36$
$α β = 3$
Therefore,
$x^{2} - (α + β) x + α β = 0$
$x^{2} - (4 x) + 3 = 0$
Hence, option $A$ is correct.

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Similar questions

Question 10
Which of the following equations has no real roots?
(A) $x^{2} - 4 x + 3 \sqrt{2} = 0$
(B) $x^{2} + 4 x - 3 \sqrt{2} = 0$
(C) $x^{2} - 4 x - 3 \sqrt{2} = 0$
(D) $3 x^{2} - 4 \sqrt{3} x + 4 = 0$

Q. If

α, β

are the root of a quadratic equation

x^{2} - 3 x + 5 = 0

, then the equation whose roots are

(α^{2} - 3 α + 7) and (β^{2} - 3 β + 7)

Q. For the quadratic equation:

x^{2} + 4 x + 5

= 0, the sum of the roots is .

Q. are the roots of quadratic equation x² - 4x + 4 = 0.

x^{2} - 4 x + 4 = 0

,does this quadratic equation has same roots?

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Which of the following quadratic equation has the sum of their roots 4 and the sum of the cubes of their roots as 28?

The correct option is A x2−4x+3=0Let α and β be the roots of the equation.Henceα+β=4and α3+β3=28Now α3+β3 can be written as=(α+β)3−3αβ(α+β)Hence28=64−12αβ12αβ=36αβ=3Therefore,x2−(α+β)x+αβ=0x2−(4x)+3=0Hence, option A is correct.

Which of the following quadratic equation has the sum of their roots $4$ and the sum of the cubes of their roots as $28$ ?