Algebra - Study Mode

[#491] If $$frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = frac{3}{2}{ ext{,}}$$ xa0xa0 then the value of $$left( {x + frac{1}{x}}
ight){ ext{is?}}$$
Correct Answer

(B) -5

Explanation

Solution: $$eqalign{
& frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = frac{3}{2}left( {{ ext{Given}}}
ight) cr
& Rightarrow frac{{xleft{ {left( {x + frac{1}{x}}
ight) - 1}
ight}}}{{xleft{ {left( {x + frac{1}{x}}
ight) + 1}
ight}}} = frac{3}{2} cr
& Rightarrow frac{{left( {x + frac{1}{x}}
ight) - 1}}{{left( {x + frac{1}{x}}
ight) + 1}} = frac{3}{2} cr
& ,,,,,,,,,,{ ext{Let }}left( {x + frac{1}{x} = y}
ight) cr
& Rightarrow frac{{y - 1}}{{y + 1}} = frac{3}{2} cr
& Rightarrow 2left( {y - 1}
ight) = 3left( {y + 1}
ight) cr
& Rightarrow 2y - 2 = 3y + 3 cr
& Rightarrow y = - 2 - 3 cr
& Rightarrow y = - 5 cr
& herefore x + frac{1}{x} = - 5 cr
& { ext{ }} cr} $$

[#492] If $$x = 3 + sqrt 8 { ext{,}}$$ xa0 then $${x^2} + frac{1}{{{x^2}}}$$ xa0 is equal to?
Correct Answer

(C) 34

Explanation

Solution: $$eqalign{
& Rightarrow x = 3 + sqrt 8 cr
& Rightarrow {x^2} = 9 + 8 + 2 imes 3sqrt 8 cr
& Rightarrow {x^2} = 17 + 6sqrt 8 cr
& Rightarrow frac{1}{{{x^2}}} = 17 - 6sqrt 8 cr
& herefore {x^2} + frac{1}{{{x^2}}} cr
& = 17 + 6sqrt 8 + 17 - 6sqrt 8 cr
& = 34 cr} $$

[#493] If $$x = 5 + 2sqrt 6 { ext{,}}$$ xa0xa0 then the value of $$left( {sqrt x + frac{1}{{sqrt x }}}
ight),{ ext{is?}}$$
Correct Answer

(C) $${ ext{2}}sqrt 3 $$

Explanation

Solution: $$eqalign{
& x = 5 + 2sqrt 6 cr
& Leftrightarrow x = 3 + 2 + 2sqrt 3 imes sqrt 2 cr
& Leftrightarrow x = {left( {sqrt 3 }
ight)^2} + {left( {sqrt 2 }
ight)^2} + 2sqrt 3 imes sqrt 2 cr
& Leftrightarrow x = {left( {sqrt 3 + sqrt 2 }
ight)^2} cr
& Leftrightarrow sqrt x = sqrt 3 + sqrt 2 cr
& { ext{Similarly,}} cr
& Leftrightarrow frac{1}{{sqrt x }} = sqrt 3 - sqrt 2 cr
& herefore left( {sqrt x + frac{1}{{sqrt x }}}
ight) cr
& = sqrt 3 + sqrt 2 + sqrt 3 - sqrt 2 cr
& = 2sqrt 3 cr} $$

[#494] If $$x + frac{9}{x} = 6{ ext{,}}$$ xa0 then $$left( {{x^2} + frac{9}{{{x^2}}}}
ight)$$ xa0 is equal to?
Correct Answer

(C) 10

Explanation

Solution: $$eqalign{
& x + frac{9}{x} = 6 cr
& { ext{Take value of x}} cr
& { ext{Let }}x = 3 cr
& 3 + frac{9}{3} = 6{ ext{ }}left( {{ ext{Proved}}}
ight) cr
& { ext{So, }}x = 3 cr
& herefore {x^2} + frac{9}{{{x^2}}} cr
& = 9 + frac{9}{9} cr
& = 10 cr
& cr
& {x08f{Alternate:}} cr
& x + frac{9}{x} = 6 cr
& { ext{On squaring,}} cr
& Rightarrow {left( {x + frac{9}{x}}
ight)^2} = 36 cr
& Rightarrow {x^2} + frac{{81}}{{{x^2}}} + 2 imes x imes frac{9}{x} = 36 cr
& Rightarrow {x^2} + frac{{81}}{{{x^2}}} - 18 = 0 cr
& Rightarrow {left( {x - frac{9}{x}}
ight)^2} = 0 cr
& Rightarrow x = frac{9}{x} cr
& Rightarrow {x^2} = 9 cr
& { ext{Hence, }}left( {{x^2} + frac{9}{{{x^2}}}}
ight) cr
& = left( {9 + frac{9}{9}}
ight) cr
& = 10 cr} $$

[#495] If $$x + frac{1}{x} = 3{ ext{,}}$$ xa0 then the value of $$frac{{{x^3} + frac{1}{x}}}{{{x^2} - x + 1}},{ ext{is?}}$$
Correct Answer

(C) $$frac{7}{2}$$

Explanation

Solution: $$eqalign{
& x + frac{1}{x} = 3{ ext{ }}left( {{ ext{Given}}}
ight) cr
& frac{{{x^3} + frac{1}{x}}}{{{x^2} - x + 1}}{ ext{ }}left( {{ ext{Divide by }}x}
ight) cr
& Rightarrow frac{{frac{{{x^3}}}{x} + frac{1}{{{x^2}}}}}{{frac{{{x^2}}}{x} - frac{x}{x} + frac{1}{x}}} cr
& Rightarrow frac{{{x^2} + frac{1}{{{x^2}}}}}{{x - 1 + frac{1}{x}}} cr
& Rightarrow frac{{{x^2} + frac{1}{{{x^2}}}}}{{x + frac{1}{x} - 1}} cr
& herefore x + frac{1}{x} = 3 cr
& herefore {x^2} + frac{1}{{{x^2}}} = 9 - 2 = 7 cr
& herefore frac{{{x^2} + frac{1}{{{x^2}}}}}{{x + frac{1}{x} - 1}} = frac{7}{{3 - 1}} = frac{7}{2} cr} $$