8 Let f(x)=(x4-7 x2+9/x-(3)x+1). If the zeros of the function f(x) are of the form (a ± √b/c) where a, b and c are positive integers then the operatornamesum( a + b + c ) is divisible by

Question

8 Let f(x)=(x4-7 x2+9/x-(3)x+1). If the zeros of the function f(x) are of the form (a ± √b/c) where a, b and c are positive integers then the operatornamesum( a + b + c ) is divisible by (A) 1 (B) 2 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

f(x)=(x[(x2-3)2-x2]/[x2+x-3])=x(x2-x-3) If f ( x )=0 ⇒ x =0 (rejected) x=(1 ± √13/2) ≡ ( a ± √ b / c ) ∴( a + b + c )=16 ⇒ 16 text is divisible by 1,2,4,8

Q. 8 Let $f (x) = \frac{x ^{4} - 7 x ^{2} + 9}{x - \frac{3}{x} + 1}$ . If the zeros of the function $f (x)$ are of the form $\frac{a \pm b}{c}$ where $a, b$ and $c$ are positive integers then the $sum (a + b + c)$ is divisible by

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$f (x) = \frac{x [ ( x ^{2} - 3 ) ^{2} - x ^{2} ]}{[ x ^{2} + x - 3 ]} = x (x^{2} - x - 3)$
If $f (x) = 0 \Rightarrow x = 0$ (rejected)
$x = \frac{1 \pm 13}{2} \equiv \frac{a \pm b}{c}$
$∴ (a + b + c) = 16 \Rightarrow 16 is divisible by 1, 2, 4, 8$

Q. 8 Let f(x)=x−x3​+1x4−7x2+9​. If the zeros of the function f(x) are of the form ca±b​​ where a,b and c are positive integers then the sum(a+b+c) is divisible by

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f(x)=[x2+x−3]x[(x2−3)2−x2]​=x(x2−x−3) If f(x)=0⇒x=0 (rejected) x=21±13​​≡ca±b​​ ∴(a+b+c)=16⇒16 is divisible by 1,2,4,8

Q. 8 Let $f (x) = \frac{x ^{4} - 7 x ^{2} + 9}{x - \frac{3}{x} + 1}$ . If the zeros of the function $f (x)$ are of the form $\frac{a \pm b}{c}$ where $a, b$ and $c$ are positive integers then the $sum (a + b + c)$ is divisible by

$f (x) = \frac{x [ ( x ^{2} - 3 ) ^{2} - x ^{2} ]}{[ x ^{2} + x - 3 ]} = x (x^{2} - x - 3)$
If $f (x) = 0 \Rightarrow x = 0$ (rejected)
$x = \frac{1 \pm 13}{2} \equiv \frac{a \pm b}{c}$
$∴ (a + b + c) = 16 \Rightarrow 16 is divisible by 1, 2, 4, 8$