Answer:
9 and 10 ques answers ....
Step-by-step explanation:
[tex]\mathrm{1.\ \ (x+y)^2-(x^2+y^2)=0}\\\mathrm{or,\ x^2+2xy+y^2-x^2-y^2=0}\\\mathrm{or,\ 2xy=0}\\\mathrm{\therefore x=0\ or\ y=0}[/tex]
[tex]\mathrm{2.\ \ (2p+q)^2-(4p^2+q^2-16)=16}\\\mathrm{or,\ \ (2p)^2+2(2p)(q)+q^2-4p^2-q^2+16=16}\\\mathrm{or,\ \ 4p^2+4pq+q^2-4p^2-q^2=16-16}\\\mathrm{or,\ \ 4pq=0}\\\mathrm{or,\ \ pq=0}\\\mathrm{\therefore p = 0\ or\ q=0}[/tex]
[tex]\mathrm{3.\ \ \frac{10x^3+5x^2+15x}{5x}=6}\\\mathrm{or,\ \ \frac{5x(2x^2+x+3)}{5x}=6,\ \ \ \ x \ne 0}\\\mathrm{or,\ \ 2x^2+x+3=6}\\\mathrm{or,\ \ 2x^2+x-3=0}\\\mathrm{or,\ \ 2x^2+(3-2)x-3=0}\\\mathrm{or,\ \ 2x^2+3x-2x-3=0}\\\mathrm{or,\ \ x(2x+3)-1(2x+3)=0}\\\mathrm{or,\ \ (2x+3)(x-1)=0}\\\mathrm{\therefore x=\frac{-3}{2}\ or\ x=1}[/tex]
[tex]\mathrm{4.\ \ \frac{4p^2q^2-12pq+9}{2pq-3}=2pq+3}\\\mathrm{or,\ \ 4p^2q^2-12pq+9=(2pq+3)(2pq-3)}\\\mathrm{or,\ \ 4p^2q^2-12pq+9=(2pq)^2-3^2}\\\mathrm{or,\ \ 4p^2q^2-12pq+9=4p^2q^2-9}\\\mathrm{or,\ \ -12pq=-9-9=-18}\\\mathrm{or,\ \ pq=\frac{3}{2}}[/tex]
[tex]\mathrm{5.\ \ y^2+y-6=(y-3)(y-2)}\\\mathrm{or,\ \ y^2+y-6=y(y-2)-3(y-2)}\\\mathrm{or,\ \ y^2+y-6=y^2-2y-3y+6}\\\mathrm{or,\ \ y^2+y-6=y^2-5y+6}\\\mathrm{or,\ \ y-6=-5y+6}\\\mathrm{or,\ 6y=12}\\\mathrm{\therefore y=2}[/tex]
[tex]\mathrm{6.\ \ 9x^2-42x+49=9x-7}\\\mathrm{or,\ \ 9x^2-42x-9x+49+7=0}\\\mathrm{or,\ \ 9x^2-51x+56=0}\\\mathrm{Using\ quadratic\ formula,}\\\mathrm{x=\frac{-(-51)\pm\sqrt{(-51)^2-4(9)(56)}}{2(9)}=\frac{51\pm3\sqrt{65}}{18}}=\frac{17\pm\sqrt{65}}{6}}[/tex]
[tex]\mathrm{7.\ \ \frac{8x^2-12x}{8x^2}=1-12x}\\\\\mathrm{or,\ \ \frac{4x(2x-3)}{4x(2x)}=1-12x,\ \ \ \ \ x\ne0}\\\\\mathrm{or,\ \ 2x-3=2x(1-12x)}\\\mathrm{or,\ \ 2x-3=2x-24x^2}\\\mathrm{or,\ \ 24x^2=3}\\\mathrm{or,\ \ x^2=\frac{1}{8}}\\\mathrm{or,\ \ x=\pm\frac{1}{2\sqrt{2}}}[/tex]
[tex]\mathrm{10. \ \ \frac{35xy}{5xy}=5xy}\\\mathrm{or,\ \ 5=5xy,\ \ xy\ne0}\\\mathrm{or,\ \ xy=1}[/tex]
I doubt that some of your questions are incorrect.
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Answer:
9 and 10 ques answers ....
Step-by-step explanation:
[tex]\mathrm{1.\ \ (x+y)^2-(x^2+y^2)=0}\\\mathrm{or,\ x^2+2xy+y^2-x^2-y^2=0}\\\mathrm{or,\ 2xy=0}\\\mathrm{\therefore x=0\ or\ y=0}[/tex]
[tex]\mathrm{2.\ \ (2p+q)^2-(4p^2+q^2-16)=16}\\\mathrm{or,\ \ (2p)^2+2(2p)(q)+q^2-4p^2-q^2+16=16}\\\mathrm{or,\ \ 4p^2+4pq+q^2-4p^2-q^2=16-16}\\\mathrm{or,\ \ 4pq=0}\\\mathrm{or,\ \ pq=0}\\\mathrm{\therefore p = 0\ or\ q=0}[/tex]
[tex]\mathrm{3.\ \ \frac{10x^3+5x^2+15x}{5x}=6}\\\mathrm{or,\ \ \frac{5x(2x^2+x+3)}{5x}=6,\ \ \ \ x \ne 0}\\\mathrm{or,\ \ 2x^2+x+3=6}\\\mathrm{or,\ \ 2x^2+x-3=0}\\\mathrm{or,\ \ 2x^2+(3-2)x-3=0}\\\mathrm{or,\ \ 2x^2+3x-2x-3=0}\\\mathrm{or,\ \ x(2x+3)-1(2x+3)=0}\\\mathrm{or,\ \ (2x+3)(x-1)=0}\\\mathrm{\therefore x=\frac{-3}{2}\ or\ x=1}[/tex]
[tex]\mathrm{4.\ \ \frac{4p^2q^2-12pq+9}{2pq-3}=2pq+3}\\\mathrm{or,\ \ 4p^2q^2-12pq+9=(2pq+3)(2pq-3)}\\\mathrm{or,\ \ 4p^2q^2-12pq+9=(2pq)^2-3^2}\\\mathrm{or,\ \ 4p^2q^2-12pq+9=4p^2q^2-9}\\\mathrm{or,\ \ -12pq=-9-9=-18}\\\mathrm{or,\ \ pq=\frac{3}{2}}[/tex]
[tex]\mathrm{5.\ \ y^2+y-6=(y-3)(y-2)}\\\mathrm{or,\ \ y^2+y-6=y(y-2)-3(y-2)}\\\mathrm{or,\ \ y^2+y-6=y^2-2y-3y+6}\\\mathrm{or,\ \ y^2+y-6=y^2-5y+6}\\\mathrm{or,\ \ y-6=-5y+6}\\\mathrm{or,\ 6y=12}\\\mathrm{\therefore y=2}[/tex]
[tex]\mathrm{6.\ \ 9x^2-42x+49=9x-7}\\\mathrm{or,\ \ 9x^2-42x-9x+49+7=0}\\\mathrm{or,\ \ 9x^2-51x+56=0}\\\mathrm{Using\ quadratic\ formula,}\\\mathrm{x=\frac{-(-51)\pm\sqrt{(-51)^2-4(9)(56)}}{2(9)}=\frac{51\pm3\sqrt{65}}{18}}=\frac{17\pm\sqrt{65}}{6}}[/tex]
[tex]\mathrm{7.\ \ \frac{8x^2-12x}{8x^2}=1-12x}\\\\\mathrm{or,\ \ \frac{4x(2x-3)}{4x(2x)}=1-12x,\ \ \ \ \ x\ne0}\\\\\mathrm{or,\ \ 2x-3=2x(1-12x)}\\\mathrm{or,\ \ 2x-3=2x-24x^2}\\\mathrm{or,\ \ 24x^2=3}\\\mathrm{or,\ \ x^2=\frac{1}{8}}\\\mathrm{or,\ \ x=\pm\frac{1}{2\sqrt{2}}}[/tex]
[tex]\mathrm{10. \ \ \frac{35xy}{5xy}=5xy}\\\mathrm{or,\ \ 5=5xy,\ \ xy\ne0}\\\mathrm{or,\ \ xy=1}[/tex]
I doubt that some of your questions are incorrect.