If -1 - -2 - -3 - -4 - -5 - -6- then the equation x - -1x - -2x - -3x - -5 - 3x - -2x - -4x - -6 - 0 has

The correct options are A Three real roots B No real root in ( ∞, α_1) f(α_1) = negative f(α_2) = zero #160; #160; #160; ∴ one root ...

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Byju's Answer

Standard VIII

Mathematics

Finding Square Root of a Number

If α1 < α2 ...

Question

If $α_{1} < α_{2} < α_{3} < α_{4} < α_{5} < α_{6}$ , then the equation $(x - α_{1}) (x - α_{2}) (x - α_{3}) (x - α_{5}) + 3 (x - α_{2}) (x - α_{4}) (x - α_{6}) = 0$ has

Three real roots

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No real root in

(- \infty, α_{1})

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One real root in

(α_{1}, α_{2})

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No real roots in

(α_{5}, α_{6})

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Solution

The correct options are
A Three real roots
B No real root in $(- \infty, α_{1})$
$f (α_{1}) =$ negative
$f (α_{2}) =$ zero $∴$ one root in $α_{2}$
$f (α_{3}) =$ positive
$f (α_{4}) =$ negative $∴$ one root in ( $α_{3}$ , $α_{4}$ )
$f (α_{5}) =$ negative
$f (α_{6}) =$ positive $∴$ one root in ( $α_{5}$ , $α_{6}$ )
Hence, there are three real roots.

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If α1<α2<α3<α4<α5<α6, then the equation (x−α1)(x−α2)(x−α3)(x−α5)+3(x−α2)(x−α4)(x−α6)=0 has

The correct options are A Three real roots B No real root in (−∞,α1)f(α1)=negativef(α2)=zero ∴ one root in α2f(α3)=positivef(α4)=negative ∴ one root in (α3, α4)f(α5)=negativef(α6)=positive ∴ one root in (α5, α6)Hence, there are three real roots.

If $α_{1} < α_{2} < α_{3} < α_{4} < α_{5} < α_{6}$ , then the equation $(x - α_{1}) (x - α_{2}) (x - α_{3}) (x - α_{5}) + 3 (x - α_{2}) (x - α_{4}) (x - α_{6}) = 0$ has