Algebra from Basic to Advance for every competitive exam CGL / CHSL / MTS / Railway etc.
⇨ $ x^4 + x^2y^2 + y^4 $
= $ (x^2)^2 + (y^2)^2 + x^2y^2 $
= $ (x^2+y^2)^2 - 2x^2y^2 + x^2y^2 $
= $ (x^2+y^2)^2 - x^2y^2 $
= $ (x^2+y^2)^2 - (xy)^2 $
= $ (x^2+xy +y^2)(x^2-xy +y^2)$
∵ [ $ a^2 - b^2 = (a+b)(a-b) $ ]
Suppose, $ x^2+xy +y^2 =$ A and $ x^2-xy +y^2 =$ B
∴ $ x^4 + x^2y^2 + y^4 = (x^2+xy +y^2)(x^2-xy +y^2) = A \times B $
⇨ $(x^2+xy +y^2) + (x^2-xy +y^2) = A + B $
or, $x^2+xy +y^2 + x^2-xy +y^2 = A + B $
or, $ 2x^2 + 2y^2 = A + B $
or, $2(x^2+y^2) = A + B $
or, $ x^2+y^2 = \cfrac{A+B}{2} $
⇨ $(x^2+xy +y^2) - (x^2-xy +y^2) = A - B $
or, $x^2+xy +y^2 - x^2+xy -y^2 = A - B $
or, $ 2xy = A - B $
or, $ xy = \cfrac{A-B}{2} $
If $ x^4 + x^2y^2 + y^4 = 189 $ and $ x^2-xy +y^2 = 9 $ , then
∴ $ x^4 + x^2y^2 + y^4 = (x^2+xy +y^2)(x^2-xy +y^2) =$ AB = 189 and $x^2-xy +y^2 = $B = 9
i. $ x^2+xy +y^2 $= A = $\cfrac{AB}{B}$$ = \cfrac{189}{9} = 21 $
ii. $ x^2 + y^2 = $$\cfrac{A+B}{2}$ $= \cfrac{21+9}{2} = \cfrac{30}{2} = 15 $
iii. $ xy = $$\cfrac{A-B}{2}$$ = \cfrac{21-9}{2} = \cfrac{12}{2} = 6 $
iv. $ (x+y)^2 = x^2 +y^2 +2xy = 15 + (2 \times 6) $
or, $ x+y =$ $\sqrt{x^2 +y^2 +2xy}$ $= \sqrt{15 + (2 \times 6)} = \sqrt{27} = 3\sqrt{3} $
v. $ (x-y)^2 = x^2 +y^2 -2xy = 15 - (2 \times 6) $
or, $ x-y = $$\sqrt{x^2 +y^2 -2xy}$ $= \sqrt{15 - (2 \times 6)} = \sqrt{3} = \sqrt{3} $
vi. $ \cfrac{x}{y}+\cfrac{y}{x} =$ $\cfrac{x^2+y^2}{xy}$$ =\cfrac{15}{6} = \cfrac{5}{2} $
vii. $ \cfrac{x^2}{y^2}+\cfrac{y^2}{x^2} = $$(\cfrac{x}{y}+\cfrac{y}{x})^2 - 2\times \cfrac{x}{y}\times\cfrac{y}{x}$ $= (\cfrac{5}{2})^2 - 2 = \cfrac{25}{4}-2 = \cfrac{25-8}{4} = \cfrac{17}{4} = 4\cfrac{1}{4} $
1. If $ x^2 - xy + y^2 = 13 $ and $ x^4 + x^2y^2 + y^4 = 312$, then the value of $ x^2 + xy + y^2 $ is
(A) 18 (B) 21 (C) 24 (D) 27
Solution
$ x^2 + xy + y^2 $
= $\cfrac{312}{13}$
= $24 $
2. If $ x^2 +y^2 = x^{-2}(777-y^4) $ and $ x+y = x^{-1}(37-y^2) $, then find the value of $ x^2+y(y-x)$ = ?
(A) 29 (B) 21 (C) 23 (D) 31
Solution
$ x^2 +y^2 = x^{-2}(777-y^4) $
or, $ x^2 +y^2 = \cfrac{1}{x^2}(777-y^4) $
or, $ x^4 +x^2y^2 + y^4 = 777 $
$ x+y = x^{-1}(37-y^2) $
or, $ x+y = \cfrac{1}{x}(37-y^2) $
or, $x^2 +xy + y^2 = 37 $
∴ $ x^2+y(y-x) $
=$ x^2 + y^2 -xy $
=$ \cfrac{777}{37} $
=$ 21$
3. If $ a+b+\sqrt{ab} = 4 $, then find $ \cfrac{a+b-\sqrt{ab}}{a^2+b^2+ab}$ =?
(A) 4 (B) 0.25 (C) 1 (D) 16
Solution
$ \cfrac{a+b-\sqrt{ab}}{a^2+b^2+ab}$
= $ \cfrac{(a+b-\sqrt{ab})}{(a+b+\sqrt{ab})(a+b-\sqrt{ab})}$
= $ \cfrac{1}{(a+b+\sqrt{ab})} $
= $ \cfrac{1}{4} $
= 0.25
4. $ x^4+x^2y^2+y^4 = 3441 $ and $ x^2-xy+y^2 = 37 $; then find $ x+y+xy = ? $
(A) 39 (B) 23 (C) 49 (D)31
Solution
$ x^2+xy+y^2 = \cfrac{3441}{37} = 93 $
∴$ x^2 +y^2 = \cfrac{93+37}{2} = 65 $
And $ xy = \cfrac{93 - 37}{2} = \cfrac{56}{2} = 28 $
Now, $ x+ y $
=$ \sqrt{65 - (2\times 28)}$
=$ \sqrt{65-56} $
=$ \sqrt{9} $
=$3 $
∴ $ x+y+xy = 3 + 28 = 31 $
5. If $ x^4+x^2y^2+y^4 = 133 $ and $ x^2-xy+y^2 = 7 $ , then find the value of $(x^2+y^2)$ is
(A) 19 (B) 16 (C) 12 (D)13
Solution
$ x^2+xy+y^2 = \cfrac{133}{7} = 19 $
∴ $ x^2 + y^2 $
=$ \cfrac{19+7}{2}$
=$ \cfrac{26}{2} = 13 $
6. If $ a^4+a^2b^2+b^4 = \cfrac{39}{512} $ and $ a^2+ab+b^2 = \cfrac{13}{32} $ , then the value of $ab$ is
(A) $ \cfrac{1}{16}$ (B)$ \cfrac{9}{64} $ (C)$ \cfrac{13}{128} $ (D)$ \cfrac{7}{64} $
Solution
$ a^2-ab+b^2 $
= $ \cfrac{39}{512} \div \cfrac{13}{32} $
=$ \cfrac{39}{512} \times \cfrac{32}{13} $
=$\cfrac{3}{16} $
∴ $ ab $
=$ \cfrac{\cfrac{13}{32} - \cfrac{3}{16}}{2} $
=$ \cfrac{\cfrac{7}{32}}{2} $
=$ \cfrac{7}{64} $
7. If $ 16x^4 + 36x^2y^2+81y^4 = 806 $ and $ 4x^2 - 6xy + 9y^2 = 26 $, then find the value of $3xy$ ?
(A) 1.5 (B) 1.25 (C)$ 1\cfrac{2}{3} $ (D) 2
Solution
$ 4x^2 + 6xy + 9y^2 $
=$ \cfrac{806}{26} $
=$31 $
∴ $ xy $
=$ \cfrac{31-26}{12} $
=$ \cfrac{5}{12} $
or, $ 3xy $
=$ 3 \times \cfrac{5}{12} $
=$ \cfrac{5}{4} $
=$ 1.25 $
8. If $ x^2 -xy +y^2 = 13$ and $ x^2 +xy+y^2 = 37$, then the value of $ \cfrac{x^6-y^6}{x^2-y^2} $ is
(A) 481 (B) 500 (C) 520 (D) 444
Solution
$ \cfrac{x^6-y^6}{x^2-y^2} $
=$ \cfrac{(x^2)^3 - (y^2)^3}{x^2-y^2} $
=$ \cfrac{(x^2 - y^2)(x^4+x^2y^2+y^4)}{(x^2-y^2)} $
=$ x^4+x^2y^2+y^4 $
= $(x^2 +xy+y^2)(x^2 -xy+y^2) $
= $37 \times 13 $
= 481
9. If $ x^4 + x^2y^2 + y^4 = \cfrac{21}{256} $ and $ x^2 +xy+y^2 = \cfrac{3}{16} $, then $(x+y)$ = ?
(A) $ \cfrac{1}{16}$ (B)$ \cfrac{5}{8} $ (C)$ \cfrac{3}{8} $ (D)$ \cfrac{1}{4} $
Solution
$ x^2 -xy+y^2 $
=$ \cfrac{21}{256} \div \cfrac{3}{16} $
=$ \cfrac{21}{256} \times \cfrac{16}{3} $
=$ \cfrac{7}{16} $
$xy $
=$ \cfrac{\cfrac{3}{16} - \cfrac{7}{16}}{2} $
=$ \cfrac{\cfrac{-4}{16}}{2} $
=$ \cfrac{-2}{16} $
$ x^2 +xy+y^2 + xy = \cfrac{3}{16} + \cfrac{-2}{16} $
or,$ x^2 +2xy+y^2 = \cfrac{3-2}{16} $
or, $ (x+y)^2 = \cfrac{1}{16} $
or, $ (x+y) = \sqrt{\cfrac{1}{16}} = \cfrac{1}{4} $
10. If $ x^2 -xy +y^2 = 17 $ and $ x^4 +x^2y^2 + y^4 = 425 $, then the value of $ \cfrac{x}{y} + \cfrac{y}{x} $ = ?
(A) 6.25 (B) 5.25 (C) 6.4 (D) 5.5
Solution
$ x^2 + xy +y^2 $
=$ \cfrac{425}{17} $
=$ 25 $
$ x^2 + y^2 $
=$ \cfrac{25+17}{2} $
=$ \cfrac{42}{2} $
=$ 21 $
$ xy $
=$ \cfrac{25-17}{2} $
=$ \cfrac{8}{2} $
=$ 4 $
$ \cfrac{x}{y} + \cfrac{y}{x} $
= $ \cfrac{x^2 + y^2 }{xy} $
= $\cfrac{21}{4} $
= 5.25
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