Algebra - Study Mode
[#321] If a 2 + a + 1 = 0, then the value of a 9 is?
Correct Answer
(C) 1
Explanation
Solution: $${a^2} + a + 1 = 0$$ [left[ x08egin{array}{l}
{a^3} + {1^3} = left( {a + 1}
ight)left( {{a^2} + a + 1}
ight)\
{a^3} - {1^3} = left( {a - 1}
ight)left( {{a^2} + a + 1}
ight)
end{array}
ight]] $$eqalign{
& herefore left( {{a^3} - 1}
ight) = left( {a - 1}
ight) imes 0 cr
& Rightarrow {a^3} - 1 = 0 cr
& Rightarrow {a^3} = 1 cr
& Rightarrow {left( {{a^3}}
ight)^3} = {1^3} cr
& Rightarrow {a^9} = 1 cr} $$
[#322] If x = -2k and y = 1 - 3k, then for what value of k, will be x = y?
Correct Answer
(B) 1
Explanation
Solution: $$eqalign{
& { ext{Given,}} cr
& x = - 2k{ ext{ and }}y = 1 - 3k cr
& herefore { ext{For }}x = y cr
& Rightarrow - 2k = 1 - 3k cr
& Rightarrow k = 1 cr} $$
[#323] If $$x + frac{1}{x} = 5{ ext{,}}$$ xa0 then $${x^6}{ ext{ + }}frac{1}{{{x^6}}}$$ xa0 is?
Correct Answer
(A) 12098
Explanation
Solution: $$eqalign{
& { ext{ }}x + frac{1}{x} = 5 cr
& { ext{Take cube on both sides}} cr
& Rightarrow {left( {x + frac{1}{x}}
ight)^3} = {left( 5
ight)^3} cr
& Rightarrow {x^3}{ ext{ + }}frac{1}{{{x^3}}} + 3 imes 5 = 125 cr
& Rightarrow {x^3}{ ext{ + }}frac{1}{{{x^3}}} = 110 cr
& herefore { ext{Squaring both sides}} cr
& Rightarrow {left( {{x^3}{ ext{ + }}frac{1}{{{x^3}}}}
ight)^2} = {left( {110}
ight)^2} cr
& Rightarrow {x^6}{ ext{ + }}frac{1}{{{x^6}}} + 2 = 12100 cr
& Rightarrow {x^6}{ ext{ + }}frac{1}{{{x^6}}} = 12100 - 2 cr
& Rightarrow {x^6}{ ext{ + }}frac{1}{{{x^6}}} = 12098 cr} $$
[#324] If x 2 - 3x + 1 = 0, then the value of $$frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}$$ xa0 xa0 will be?
Correct Answer
(C) 21
Explanation
Solution: $$eqalign{
& {x^2} - 3x + 1 = 0 cr
& Rightarrow {x^2} + 1 = 3x cr
& Rightarrow x + frac{1}{x} = 3 cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} + 3 imes 3 = 27 cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} = 18 cr
& herefore frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}{ ext{ }} cr
& = frac{{{x^6}}}{{{x^3}}} + frac{{{x^4}}}{{{x^3}}} + frac{{{x^2}}}{{{x^3}}} + frac{1}{{{x^3}}} cr
& = {x^3} + frac{1}{{{x^3}}} + frac{1}{x} + x cr
& = 18 + 3 cr
& = 21 cr} $$
[#325] If $$frac{p}{a}$$ + $$frac{q}{b}$$ + $$frac{r}{c}$$ = 1 and $$frac{a}{p}$$ + $$frac{b}{q}$$ + $$frac{c}{r}$$ = 0 where p, q, r and a, b, c are non - zero, then value of $$frac{{{p^2}}}{{{a^2}}}$$ + $$frac{{{q^2}}}{{{b^2}}}$$ + $$frac{{{r^2}}}{{{c^2}}}$$ = ?
Correct Answer
(C) 1
Explanation
Solution: $$eqalign{
& frac{p}{a} + frac{q}{b} + frac{r}{c} = 1 cr
& frac{a}{p} + frac{b}{q} + frac{c}{r} = 0{ ext{ }} cr
& Rightarrow frac{p}{a} = x,{ ext{ }}frac{q}{b} = y,{ ext{ }}frac{r}{c} = z cr
& Rightarrow left( {x + y + z}
ight) = 1 cr
& { ext{Squaring the both sides}} cr
& Rightarrow {x^2} + {y^2} + {z^2} + 2left( {xy + yz + zx}
ight) = 1 cr
& { ext{and }}frac{1}{x} + frac{1}{y} + frac{1}{z} = 0 cr
& Rightarrow frac{{xy + yz + zx}}{{xyz}} = 0 cr
& Rightarrow xy + yz + zx = 0 cr
& herefore {x^2} + {y^2} + {z^2} = 1 cr
& { ext{So, }}frac{{{p^2}}}{{{a^2}}} + frac{{{q^2}}}{{{b^2}}} + frac{{{r^2}}}{{{c^2}}} = 1 cr} $$