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Question : 26 of 39

Marks: +1, -0

If

\alpha, \beta

are the roots of equation

x^{2}-5x+6=0

\alpha>\beta

then the equation with roots

\alpha^{2}+\beta)

(\alpha+\beta^{2})

is

$x^{2}-9x+99=0$
$x^{2}-18x+90=0$
$x^{2}-18x+77=0$
None

Solution:

Concept:

Given a quadratic and order of roots, compute the new roots and form the quadratic.

Explanation:

Roots of

x^{2}-5x+6=0

are

(x-2)(x-3)=0

, so

x=2,3

.

Given

\alpha>\beta

, so

\alpha=3

,

\beta=2

.

First new root:

\alpha^{2}+\beta = 3^{2}+2 = 9+2=11

.

Second new root:

\alpha+\beta^{2} = 3+2^{2} = 3+4=7

.

Equation with roots

11

and

7

:

x^{2} - (11+7)x + (11\cdot7)=0

.

That is

x^{2}-18x+77=0

.

Answer:

x^{2}-18x+77=0

(Option C)

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