Class 10 Math All topics Quadratic Equations

Solve $x+2+y+3+(x+2)(y+3) =39$.

$(x+2)_{2}+(y+3)_{2}+(x+2)(y+3)=741$.

$(x+2)_{2}+(y+3)_{2}+(x+2)(y+3)=741$.

Solution: $x+2+y+3+(x+2)(y+3) =39$

$(x+2)_{2}+(y+3)_{2}+(x+2)(y+3)=741$

Let $x+2=p,y+3=q.$

$p+q+pq =39⟶I$

$p_{2}+q_{2}+pq=741⟶II$

$⟹p+q−39=−pq $

Squaring on both sides:-

$⟹p_{2}+q_{2}+1521+2pq−78p−78q=pq$

$⟹p_{2}+q_{2}+pq=78p+78q−1521⟶III$

Subtracting $II$ from $III$

$78p+78q−1521−741=0$

$78p+78q−2262=0$

$p+q=29⟶IV$

From $I$

$pq =10$

$pq=100⟶V$

$p−q=(p+q)_{2}−4pq $

$=841−400 $

$=441 =21⟶VI$

Adding $IV$ and $VI$

$2p=50$

$p=25,q=4$

$x+2=25,y+3=4$

$x=23,y=1 $

$(x+2)_{2}+(y+3)_{2}+(x+2)(y+3)=741$

Let $x+2=p,y+3=q.$

$p+q+pq =39⟶I$

$p_{2}+q_{2}+pq=741⟶II$

$⟹p+q−39=−pq $

Squaring on both sides:-

$⟹p_{2}+q_{2}+1521+2pq−78p−78q=pq$

$⟹p_{2}+q_{2}+pq=78p+78q−1521⟶III$

Subtracting $II$ from $III$

$78p+78q−1521−741=0$

$78p+78q−2262=0$

$p+q=29⟶IV$

From $I$

$pq =10$

$pq=100⟶V$

$p−q=(p+q)_{2}−4pq $

$=841−400 $

$=441 =21⟶VI$

Adding $IV$ and $VI$

$2p=50$

$p=25,q=4$

$x+2=25,y+3=4$

$x=23,y=1 $

Similar topics

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

polynomials

introduction to trigonometry

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