SEBA Class 10 Advanced Maths Chapter 4.3 Solutions: Quadratic Equation
Get free SEBA Class 10 Advanced Maths Chapter 4.3 Solutions for the upcoming HSLC exam held in Assam. In this article, we solved all the textual exercises of 4.3 from the SEBA Class 10 Advanced Mathematics book. We provide all solutions of Chapter 4.2 in a very simple way.
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SEBA Class 10 Advanced Maths Chapter 4.3 Solutions
Exercise 4.3 Solutions
Q1. Solve the following equations:
$3x-2y=2, 9x^{2}+4y^{2}=2$
Soln :
$3x-2y=2$ ….. (i)
$9x^{2}+4y^{2}=2$ ….. (ii)
From (i) $\Rightarrow 3x=2+2y$ ….. (iii)
Putting $3x=2+2y$ in equation (ii), we get,
$(2+2y)^{2}+4y^{2}=2$
$\Rightarrow 4+8y+4y^{2}+4y^{2}-2=0$
$\Rightarrow 8y^{2}+8y+2=0$
$\Rightarrow 2(4y^{2}+4y+1)=0$
$\Rightarrow 4y^{2}+4y+1=0$
$\Rightarrow (2y+1)^{2}=0$
$\Rightarrow (2y+1)(2y+1)=0$
Either,
$2y+1=0$
$\Rightarrow y=-\frac{1}{2}$
Or,
$2y+1=0$
$\Rightarrow y=-\frac{1}{2}$
When $y=-\frac{1}{2}$, (iii) $\Rightarrow 3x=2+2(-\frac{1}{2})$
$\Rightarrow 3x=2-1$
$\Rightarrow x=\frac{1}{3}$
Again, when $y=-\frac{1}{2}$, (iii) $\Rightarrow 3x=2+2(-\frac{1}{2})=2-1$
$\Rightarrow x=\frac{1}{3}$
$\therefore$ Required roots $(\frac{1}{3},-\frac{1}{2}), (\frac{1}{3},-\frac{1}{2})$
2. $x-y+2=0, x^{2}+y^{2}=100$
Soln :
$x-y+2=0$ ….. (i)
$x^{2}+y^{2}=100$ ….. (ii)
From (i) $\Rightarrow x=y-2$ ….. (iii)
Putting $x=y-2$ in equation (ii), we get,
$(y-2)^{2}+y^{2}=100$
$\Rightarrow y^{2}-4y+4+y^{2}-100=0$
$\Rightarrow 2y^{2}-4y-96=0$
$\Rightarrow 2(y^{2}-2y-48)=0$
$\Rightarrow y^{2}-2y-48=0$
$\Rightarrow y^{2}-(8-6)y-48=0$
$\Rightarrow y^{2}-8y+6y-48=0$
$\Rightarrow y(y-8)+6(y-8)=0$
$\Rightarrow (y-8)(y+6)=0$
Either,
$y-8=0$
$\Rightarrow y=8$
Or,
$y+6=0$
$\Rightarrow y=-6$
When $y=8$, (iii) $\Rightarrow x=8-2=6$
When $y=-6$, (iii) $\Rightarrow x=-6-2=-8$
$\therefore$ Required roots $(6,8)$ and $(-8,-6)$.
3. Solve: $x+y=8, x^{2}+y^{2}=50$
Soln :
$x+y=8$ ….. (i)
$x^{2}+y^{2}=50$ ….. (ii)
From (i) $\Rightarrow x=8-y$ ….. (iii)
Putting $x=8-y$ in equation (ii), we get,
$(8-y)^{2}+y^{2}=50$
$\Rightarrow 64-16y+y^{2}+y^{2}=50$
$\Rightarrow 2y^{2}-16y+14=0$
$\Rightarrow 2(y^{2}-8y+7)=0$
$\Rightarrow y^{2}-8y+7=0$
$\Rightarrow y^{2}-(7+1)y+7=0$
$\Rightarrow y^{2}-7y-y+7=0$
$\Rightarrow y(y-7)-1(y-7)=0$
$\Rightarrow (y-7)(y-1)=0$
Either,
$y-7=0$
$\Rightarrow y=7$
Or,
$y-1=0$
$\Rightarrow y=1$
When $y=7$, (iii) $\Rightarrow x=8-7=1$
When $y=1$, (iii) $\Rightarrow x=8-1=7$
$\therefore$ Required roots $(1,7)$ and $(7,1)$
4 Solve: $x+3y+2=0, 4x^{2}+3y^{2}=7$
Soln :
$x+3y+2=0$ ….. (i)
$4x^{2}+3y^{2}=7$ ….. (ii)
From (i), we get $x=-3y-2 = -(3y+2)$ ….. (iii)
Putting $x=-(3y+2)$ in equation (ii), we get,
$4\{-(3y+2)\}^{2}+3y^{2}=7$
$\Rightarrow 4(9y^{2}+12y+4)+3y^{2}=7$
$\Rightarrow 36y^{2}+48y+16+3y^{2}-7=0$
$\Rightarrow 39y^{2}+48y+9=0$
$\Rightarrow 3(13y^{2}+16y+3)=0$
$\Rightarrow 13y^{2}+16y+3=0$
$\Rightarrow 13y^{2}+13y+3y+3=0$
$\Rightarrow 13y(y+1)+3(y+1)=0$
$\Rightarrow (y+1)(13y+3)=0$
Either,
$y+1=0$
$\Rightarrow y=-1$
Or,
$13y+3=0$
$\Rightarrow y=-\frac{3}{13}$
When $y=-1$, (iii) $\Rightarrow x=-3(-1)-2 = 3-2 = 1$
When $y=-\frac{3}{13}$, (iii)
$\Rightarrow x=-3(-\frac{3}{13})-2 $
$\Rightarrow x = \frac{9}{13}-2$
$\Rightarrow x= -\frac{17}{13}$
$\therefore$ Required Solution : $(1,-1)$ and $(-\frac{17}{13},-\frac{3}{13})$
5. Solve: $x-3y=1, x^{2}-5xy+2y^{2}+2=0$
Soln :
$x-3y=1$ ….. (i)
$x^{2}-5xy+2y^{2}+2=0$ ….. (ii)
From (i), we get, $x=1+3y$ ….. (iii)
Putting $x=1+3y$ in equation (ii), we get,
$(1+3y)^{2}-5(1+3y)y+2y^{2}+2=0$
$\Rightarrow 1+6y+9y^{2}-5y-15y^{2}+2y^{2}+2=0$
$\Rightarrow -4y^{2}+y+3=0$
$\Rightarrow 4y^{2}-y-3=0$
$\Rightarrow 4y^{2}-4y+3y-3=0$
$\Rightarrow 4y(y-1)+3(y-1)=0$
$\Rightarrow (y-1)(4y+3)=0$
Either,
$y-1=0$
$\Rightarrow y=1$
Or,
$4y+3=0$
$\Rightarrow y=-\frac{3}{4}$
Putting $y=1$ in (iii), we get
$x=1+3(1)=4$
When $y=-\frac{3}{4}$ in (iii), we get
$\Rightarrow x=1+3(-\frac{3}{4})$
$\Rightarrow x=1-\frac{9}{4} $
$\Rightarrow x= -\frac{5}{4}$
$\therefore$ Required Solution : $(4,1)$ and $(-\frac{5}{4},-\frac{3}{4})$
6. Solve: $x-y=2, x^{2}-3xy-2y^{2}-2=0$
Soln :
$x-y=2$ ….. (i)
$x^{2}-3xy-2y^{2}-2=0$ ….. (ii)
From (i), we get $x=2+y$ ….. (iii)
Putting $x=2+y$ in equation (ii), we get,
$(2+y)^{2}-3(2+y)y-2y^{2}-2=0$
$\Rightarrow 4+4y+y^{2}-6y-3y^{2}-2y^{2}-2=0$
$\Rightarrow -4y^{2}-2y+2=0$
$\Rightarrow -2(2y^{2}+y-1)=0$
$\Rightarrow 2y^{2}+y-1=0$
$\Rightarrow 2y^{2}+(2-1)y-1=0$
$\Rightarrow 2y^{2}+2y-y-1=0$
$\Rightarrow 2y(y+1)-1(y+1)=0$
$\Rightarrow (y+1)(2y-1)=0$
Either,
$y+1=0$
$\Rightarrow y=-1$
Or,
$2y-1=0$
$\Rightarrow y=\frac{1}{2}$
When $y=-1$, (iii)
$\Rightarrow x=2-1=1$
When $y=\frac{1}{2}$, (iii)
$\Rightarrow x=2+\frac{1}{2}$
$\Rightarrow x=\frac{5}{2}$
$\therefore$ Required roots: $(1,-1)$ and $(\frac{5}{2},\frac{1}{2})$
7. Solve: $x+3y=7, 2x^{2}+3xy+4y^{2}=24$
Soln :
$x+3y=7$ ….. (i)
$2x^{2}+3xy+4y^{2}=24$ ….. (ii)
From (i), $x=7-3y$ ….. (iii)
Putting $x=7-3y$ in (ii), we get,
$2(7-3y)^{2}+3(7-3y)y+4y^{2}=24$
$\Rightarrow 2(49-42y+9y^{2})+21y-9y^{2}+4y^{2}=24$
$\Rightarrow 98-84y+18y^{2}+21y-5y^{2}-24=0$
$\Rightarrow 13y^{2}-63y+74=0$
$\Rightarrow 13y^{2}-37y-26y+74=0$
$\Rightarrow y(13y-37)-2(13y-37)=0$
$\Rightarrow (13y-37)(y-2)=0$
Either,
$13y-37=0$
$\Rightarrow y=\frac{37}{13}$
Or,
$y-2=0$
$\Rightarrow y=2$
When $y=\frac{37}{13}$, putting in (iii), we get
$x=7-3 \times \frac{37}{13} $
$x= \frac{91-111}{13} $
$x= -\frac{20}{13}$
When $y=2$, (iii)
$\Rightarrow x=7-3 \times 2 = 1$
$\Rightarrow x=7-6 = 1$
$\therefore$ Required Solution : $(1,2)$ and $(-\frac{20}{13},\frac{37}{13})$
8. Solve: $x+y+\sqrt{xy}=28, x^{2}+y^{2}+xy=336$
Soln :
$x+y+\sqrt{xy}=28$ ….. (i)
$x^{2}+y^{2}+xy=336$ ….. (ii)
From (ii),
$x^{2}+y^{2}+xy=336$
$\Rightarrow (x+y)^{2}-2xy+xy=336$
$\Rightarrow (x+y)^{2}-xy=336$
$\Rightarrow (x+y)^{2}-(\sqrt{xy})^{2}=336$
$\Rightarrow (x+y+\sqrt{xy})(x+y-\sqrt{xy})=336$
$\Rightarrow 28 \times (x+y-\sqrt{xy})=336$ [Using (i)]
$\Rightarrow x+y-\sqrt{xy}=12$ ….. (iii)
Adding (i) and (iii), we get $2(x+y)=40$
$\Rightarrow x+y=20$ $\Rightarrow x=20-y$ ….. (iv)
Putting $x=20-y$ in (ii), we get,
$(20-y)^{2}+y^{2}+(20-y)y=336$
$\Rightarrow 400-40y+y^{2}+y^{2}+20y-y^{2}=336$
$\Rightarrow y^{2}-20y+64=0$
$\Rightarrow y^{2}-16y-4y+64=0$
$\Rightarrow y(y-16)-4(y-16)=0$
$\Rightarrow (y-4)(y-16)=0$
Either, $y-4=0$ or $y-16=0$
$\Rightarrow y=4$ or $ y=16$
Putting $y=4$ in (iv), we get
$x=20-4=16$
Again, when $y=16$, in (iv) we get
$x=20-16=4$
$\therefore$ Required Solution: $(16,4)$ and $(4,16)$
9. Solve: $x+y-\sqrt{xy}=6, x^{2}+y^{2}+xy=84$
Soln :
$x+y-\sqrt{xy}=6$ ….. (i)
$x^{2}+y^{2}+xy=84$ ….. (ii)
From (ii),
$x^{2}+y^{2}+xy=84$
$\Rightarrow (x^2 + 2xy + y^2) – xy = 84$
$\Rightarrow (x + y)^2 – (\sqrt{xy})^2 = 84$
$\Rightarrow (x+y+\sqrt{xy})(x+y-\sqrt{xy})=84$
$\Rightarrow (x+y+\sqrt{xy}) \cdot 6 = 84$ [From eq(i)]
$\Rightarrow x+y+\sqrt{xy}=14$ ….. (iii)
Adding (i) and (iii), we get
$2(x+y)=20$
$\Rightarrow x+y=10$
$\Rightarrow x=10-y$ ….. (iv)
Putting $x=10-y$ in (ii), we get,
$(10-y)^{2}+y^{2}+(10-y)y=84$
$\Rightarrow 100-20y+y^{2}+y^{2}+10y-y^{2}=84$
$\Rightarrow y^{2}-10y+16=0$
$\Rightarrow y^{2}-8y-2y+16=0$
$\Rightarrow y(y-8)-2(y-8)=0$
$\Rightarrow (y-8)(y-2)=0$
Either, $y-8=0$ or $y-2=0$
$\Rightarrow y=8$ or $ y=2$
Putting $y=8$ in (iv), we get
$x=10-8=2$
Again, when $y=2$, in (iv) we get
$x=10-2=8$
$\therefore$ Required Solution: $(2,8)$ and $(8,2)$
10. Solve: $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{10}{3}, x+y=10$
Soln .
$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{10}{3}$ ….. (i)
$x+y=10$ ….. (ii)
From (i), we get
$\Rightarrow \frac{x+y}{\sqrt{xy}}=\frac{10}{3}$
$\Rightarrow \frac{10}{\sqrt{xy}}=\frac{10}{3}$ [From eq(ii)]
$\Rightarrow \sqrt{xy}=3$ ….. (iii)
From (ii), we get $x=10-y$ ….. (iv)
Putting $x=10-y$ in (iii), we get,
$\sqrt{(10-y)y}=3$
$\Rightarrow 10y-y^{2}=9$ (Squaring both sides)
$\Rightarrow -y^{2}+10y-9=0$
$\Rightarrow y^{2}-10y+9=0$
$\Rightarrow y^{2}-9y-y+9=0$
$\Rightarrow y(y-9)-1(y-9)=0$
$\Rightarrow (y-9)(y-1)=0$
Either,
$y-9=0$ or $y-1=0$
$\Rightarrow y=9$ or $y=1$
Putting $y=9$ in (iv), we get
$x=10-9=1$
Again, when $y=1$, (iv)
$\Rightarrow x=10-1=9$
$\therefore$ Required Solution: $(1,9)$ and $(9,1)$
11. Solve: $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{5}{2}, x+y=20$
Soln .
$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{5}{2}$ ….. (i)
$x+y=20$ ….. (ii)
From (i), we get;
$\Rightarrow \frac{x+y}{\sqrt{xy}}=\frac{5}{2}$
$\Rightarrow \frac{20}{\sqrt{xy}}=\frac{5}{2}$ [from eq(ii)]
$\Rightarrow \sqrt{xy}=8$ ….. (iii)
From (ii), we get, $x=20-y$ ….. (iv)
Putting $x=20-y$ in (iii), we get,
$\sqrt{(20-y)y}=8$
$\Rightarrow 20y-y^{2}=64$ (Squaring both sides we get)
$\Rightarrow -y^{2}+20y-64=0$
$\Rightarrow -(y^{2}-20y+64)=0$
$\Rightarrow y^{2}-20y+64=0$
$\Rightarrow y^{2}-16y-4y+64=0$
$\Rightarrow y(y-16)-4(y-16)=0$
$\Rightarrow (y-16)(y-4)=0$
Either,
$y-16=0$ or $y-4=0$
$\Rightarrow y=16$ or $ y=4$
When $y=16$, (iv) $\Rightarrow x=20-16=4$
When $y=4$, (iv) $\Rightarrow x=20-4=16$
$\therefore$ Required Solution: $(4,16)$ and $(16,4)$
12. Solve: $x+y=p+q, \frac{p}{x}+\frac{q}{y}=2$
Soln .
$x+y=p+q$ ….. (i)
$\frac{p}{x}+\frac{q}{y}=2$ ….. (ii)
From (ii), we get
$\frac{p}{x}+\frac{q}{y}=2$
$\Rightarrow \frac{p}{x}-1+\frac{q}{y}-1=0$
$\Rightarrow \frac{p-x}{x}+\frac{q-y}{y}=0$
$\Rightarrow \frac{y-q}{x}-\frac{y-q}{y}=0$ $[\because x+y=p+q \Rightarrow y-q=p-x]$
$\Rightarrow (y-q)(\frac{1}{x}-\frac{1}{y})=0$
Either,
$y-q=0$
$\Rightarrow y=q$
Or,
$\frac{1}{x}-\frac{1}{y}=0$
$\Rightarrow \frac{1}{x}=\frac{1}{y} $
$\Rightarrow x=y$
When $y=q$, (i) $\Rightarrow x+q=p+q $
$\Rightarrow x=p$
When $x=y$, (i) $\Rightarrow x+x=p+q$
$\Rightarrow 2x=p+q $
$\Rightarrow x=\frac{p+q}{2}$
$\therefore y=\frac{p+q}{2}$ (as $x=y$)
$\therefore$ Required Solution: $(p,q)$ and $(\frac{p+q}{2},\frac{p+q}{2})$
13. Solve: $\frac{1}{x}+\frac{1}{y}=\frac{3}{2}, \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{5}{4}$
Sol.
$\frac{1}{x}+\frac{1}{y}=\frac{3}{2}$ ….. (i)
$\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{5}{4}$ ….. (ii)
From (ii),
$\Rightarrow (\frac{1}{x}+\frac{1}{y})^{2}-\frac{2}{xy}=\frac{5}{4}$
$\Rightarrow (\frac{3}{2})^{2}-\frac{2}{xy}=\frac{5}{4}$ [ from eq (i) ]
$\Rightarrow \frac{9}{4}-\frac{2}{xy}=\frac{5}{4}$
$\Rightarrow -\frac{2}{xy}=\frac{5}{4}-\frac{9}{4}$
$\Rightarrow -\frac{2}{xy}=-1$
$\Rightarrow \frac{2}{xy}=1$
$\Rightarrow \frac{1}{x}=\frac{y}{2}$ ….. (iii)
Putting $\frac{1}{x}=\frac{y}{2}$ in (i), we get,
$\frac{y}{2}+\frac{1}{y}=\frac{3}{2}$
$\Rightarrow \frac{y^{2}+2}{2y}=\frac{3}{2}$
$\Rightarrow 2y^{2}+4=6y$
$\Rightarrow 2y^{2}-6y+4=0$
$\Rightarrow 2(y^{2}-3y+2)=0$
$\Rightarrow y^{2}-3y+2=0$
$\Rightarrow y^{2}-2y-y+2=0$
$\Rightarrow y(y-2)-1(y-2)=0$
$\Rightarrow (y-2)(y-1)=0$
Either $y-2=0$ or $y-1=0$
$\Rightarrow y=2$ or $y=1$
When $y=2$, (iii) $\Rightarrow \frac{1}{x}=1 \Rightarrow x=1$
Again, when $y=1$, (iii) $\Rightarrow \frac{1}{x}=\frac{1}{2} \Rightarrow x=2$
$\therefore$ Required roots: $(1,2)$ and $(2,1)$
14. Solve: $\frac{a}{x}+\frac{b}{y}=2, \frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=2$
Sol.
$\frac{a}{x}+\frac{b}{y}=2$ ….. (i)
$\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=2$ ….. (ii)
Again, from (ii) we get,
$\Rightarrow (\frac{a}{x}+\frac{b}{y})^{2}-\frac{2ab}{xy}=2$
$\Rightarrow 2^{2}-\frac{2ab}{xy}=2$
$\Rightarrow -\frac{2ab}{xy}=-2$
$\Rightarrow \frac{ab}{xy}=1$
$\Rightarrow \frac{b}{y}=\frac{x}{a}$ ….. (iii)
Putting $\frac{b}{y}=\frac{x}{a}$ in (i), we get,
$\frac{a}{x}+\frac{x}{a}=2$
$\Rightarrow \frac{a^{2}+x^{2}}{xa}=2$
$\Rightarrow a^{2}+x^{2}=2xa$
$\Rightarrow x^{2}-2xa+a^{2}=0$
$\Rightarrow (x-a)^{2}=0$
$\Rightarrow (x-a)(x-a)=0$
$\Rightarrow x=a, a$
When $x=a$, (iii) $\Rightarrow \frac{b}{y}=1 \Rightarrow y=b$
Again, when $x=a$, (iii) $\Rightarrow y=b$
$\therefore$ Required roots: $(a,b)$ and $(a,b)$
15. Solve: $\frac{4x}{3}-\frac{6y}{5}=1, \frac{16x^{2}}{9}-\frac{36y^{2}}{25}=3$
Sol.
$\frac{4x}{3}-\frac{6y}{5}=1$ ….. (i)
$\frac{16x^{2}}{9}-\frac{36y^{2}}{25}=3$ ….. (ii)
From (ii) we get,
$\Rightarrow (\frac{4x}{3}-\frac{6y}{5})(\frac{4x}{3}+\frac{6y}{5})=3$
$\Rightarrow 1 \cdot (\frac{4x}{3}+\frac{6y}{5})=3$ [from eq(i)]
$\Rightarrow \frac{4x}{3}+\frac{6y}{5}=3$ ….. (iii)
(i) + (iii) $\Rightarrow 2 \times \frac{4x}{3}=4$
$\Rightarrow \frac{2x}{3}=1$
$\Rightarrow x=\frac{3}{2}$
Putting $x=\frac{3}{2}$ in (i), we get,
$\frac{4}{3} \times \frac{3}{2}-\frac{6y}{5}=1$
$\Rightarrow 2-\frac{6y}{5}=1$
$\Rightarrow \frac{6y}{5}=1$
$\Rightarrow y=\frac{5}{6}$
$\therefore$ Required Roots: $(\frac{3}{2},\frac{5}{6})$
16. Solve: $8 \cdot 2^{xy}=4^{y}, 27 \cdot 9^{x} \cdot 3^{xy}=1$
Sol.
$8 \cdot 2^{xy}=4^{y}$ ….. (i)
$27 \cdot 9^{x} \cdot 3^{xy}=1$ ….. (ii)
From (i), we get,
$\Rightarrow 2^{3} \cdot 2^{xy}=2^{2y}$
$\Rightarrow 2^{3+xy}=2^{2y}$
$\Rightarrow 3+xy=2y $
$\Rightarrow xy=2y-3$ ….. (iii)
From (ii), we get,
$\Rightarrow 3^{3} \cdot 3^{2x} \cdot 3^{xy}=1$
$\Rightarrow 3^{3+2x+xy}=3^{0}$
$\therefore 3+2x+xy=0$ ….. (iv)
Putting $xy=2y-3$ in (iv), we get,
$3+2x+2y-3=0$
$\Rightarrow 2(x+y)=0$
$\Rightarrow x+y=0$
$\Rightarrow x=-y$ ….. (v)
Putting $x=-y$ in (iii), we get,
$-y^{2}=2y-3$
$\Rightarrow y^{2}+2y-3=0$
$\Rightarrow y^{2}+3y-y-3=0$
$\Rightarrow y(y+3)-1(y+3)=0$
$\Rightarrow (y+3)(y-1)=0$
Either $y + 3 =0$ or $y-1=0$
$\Rightarrow y=-3$ or $y=1$
When $y=-3$, (v) $\Rightarrow x=3$
When $y=1$, (v) $\Rightarrow x=-1$
∴$ Required Roots: $(-1,3)$ and $(1,-3)$
17 Solve: $3^{x}=9^{y}, 5^{x+y+1}=25^{xy}$
Sol.
$3^{x}=9^{y}$
$\Rightarrow 3^{x}=3^{2y}$
$\therefore x=2y$ ….. (i)
$5^{x+y+1}=25^{xy}$
$\Rightarrow 5^{x+y+1}=5^{2xy}$
$\therefore x+y+1=2xy$ ….. (ii)
Putting $x=2y$ in (ii), we get
$2y+y+1=2(2y)y$
$\Rightarrow 3y+1=4y^{2}$
$\Rightarrow 4y^{2}-3y-1=0$
$\Rightarrow 4y^{2}-(4-1)y-1=0$
$\Rightarrow 4y^{2}-4y+y-1=0$
$\Rightarrow 4y(y-1)+1(y-1)=0$
$\Rightarrow (y-1)(4y+1)=0$
Either $y-1=0$ or $4y+1=0$
$\Rightarrow y=1$ or $ y=-\frac{1}{4}$
When $y=1$, (i) $\Rightarrow x=2$
When $y=-\frac{1}{4}$, (i) $\Rightarrow x=-\frac{1}{2}$
∴$ Required Roots: $(2,1)$ and $(-\frac{1}{2},-\frac{1}{4})$
18. Solve: $8^{xy}=2^{x-1}, 3^{y+8}=9^{5y+x}$
Sol.
$8^{xy}=2^{x-1}$
$\Rightarrow 2^{3xy}=2^{x-1}$
$\therefore 3xy=x-1$ ….. (i)
Again, $3^{y+8}=9^{5y+x}$
$\Rightarrow 3^{y+8}=3^{10y+2x}$
$\therefore y+8=10y+2x$
$\Rightarrow 8-2x=9y$
$\Rightarrow y=\frac{8-2x}{9}$ ….. (ii)
Putting $y=\frac{8-2x}{9}$ in (i), we get
$3x(\frac{8-2x}{9})=x-1$
$\Rightarrow \frac{8x-2x^{2}}{3}=x-1$
$\Rightarrow 8x-2x^{2}=3x-3$
$\Rightarrow -2x^{2}+5x+3=0$
$\Rightarrow 2x^{2}-5x-3=0$
$\Rightarrow 2x^{2}-6x+x-3=0$
$\Rightarrow 2x(x-3)+1(x-3)=0$
$\Rightarrow (x-3)(2x+1)=0$
Either $x-3=0$ or $2x+1=0$
$\Rightarrow x=3$ or $ x=-\frac{1}{2}$
When $x=3$, (ii) $\Rightarrow y=\frac{8-2(3)}{9}=\frac{2}{9}$
Again, when $x=-\frac{1}{2}$, (ii) $\Rightarrow y=\frac{8-2(-\frac{1}{2})}{9} $
$\Rightarrow y=\frac{9}{9}=1$
$\therefore$ Required roots: $(3,\frac{2}{9})$ and $(-\frac{1}{2},1)$
19. Solve: $x^{\frac{1}{3}}+y^{\frac{1}{3}}=3, x+y=9$
Sol.
$x^{\frac{1}{3}}+y^{\frac{1}{3}}=3$ …..
(i) $x+y=9$ ….. (ii)
Now, from (i)
$\Rightarrow x^{\frac{1}{3}}+y^{\frac{1}{3}}=3$
$\Rightarrow (x^{\frac{1}{3}}+y^{\frac{1}{3}})^{3}=3^{3}$ (Cubing both sides we get)
$\Rightarrow (x^{\frac{1}{3}})^{3}+(y^{\frac{1}{3}})^{3}+3 \cdot x^{\frac{1}{3}} \cdot y^{\frac{1}{3}}(x^{\frac{1}{3}}+y^{\frac{1}{3}})=27$
$\Rightarrow x+y+3 \cdot x^{\frac{1}{3}} \cdot y^{\frac{1}{3}} \cdot 3=27$ [from eq(i)]
$\Rightarrow x+y+9x^{\frac{1}{3}}y^{\frac{1}{3}}=27$
$\Rightarrow 9+9x^{\frac{1}{3}}y^{\frac{1}{3}}=27$ [From eq(ii)]
$\Rightarrow 9(1+x^{\frac{1}{3}}y^{\frac{1}{3}})=27$
$\Rightarrow 1+x^{\frac{1}{3}}y^{\frac{1}{3}}=3$ $\Rightarrow x^{\frac{1}{3}}y^{\frac{1}{3}}=2$
$\Rightarrow xy=8$ ….. (iii)
From (ii), we get $x=9-y$ ….. (iv)
Putting $x=9-y$ in (iii), we get
$(9-y)y=8$
$\Rightarrow 9y-y^{2}=8$
$\Rightarrow -y^{2}+9y-8=0$
$\Rightarrow -(y^{2}-9y+8)=0$
$\Rightarrow y^{2}-9y+8=0$
$\Rightarrow y^{2}-(8+1)y+8=0$
$\Rightarrow y^{2}-8y-y+8=0$
$\Rightarrow y(y-8)-1(y-8)=0$
$\Rightarrow (y-8)(y-1)=0$
Either $y-8=0$ or $y-1=0$
$\Rightarrow y=8$ or $y=1$
When $y=8$, (iv) $\Rightarrow x=9-8=1$
When $y=1$, (iv) $\Rightarrow x=9-1=8$
$\therefore $ Required roots: $(1,8)$ and $(8,1)$
20. Find the co-ordinates of the points at which x-axis and y-axis intersect the curves represented by the following equations:
(i) $2x^{2}-3xy+y^{2}+x-2y-3=0$
Sol.
$2x^{2}-3xy+y^{2}+x-2y-3=0$ ….. (i)
At x-axis, $y=0$
$\therefore $ When $y=0$, (i) $\Rightarrow 2x^{2}+x-3=0$
$\Rightarrow 2x^{2}+3x-2x-3=0$
$\Rightarrow x(2x+3)-1(2x+3)=0$
$\Rightarrow (2x+3)(x-1)=0$
Either,
$2x+3=0 $
$\Rightarrow 2x=-3 $
$\Rightarrow x=-\frac{3}{2}$
Or,
$x-1=0 $
$\Rightarrow x=1$
Hence the curve intersects x-axis at $(-\frac{3}{2},0)$ and $(1,0)$.
At y-axis, $x=0$
When $x=0$, (i) $\Rightarrow y^{2}-2y-3=0$
$\Rightarrow y^{2}-(3-1)y-3=0$
$\Rightarrow y^{2}-3y+y-3=0$
$\Rightarrow y(y-3)+1(y-3)=0$
$\Rightarrow (y-3)(y+1)=0$
Either,
$y-3=0 \Rightarrow y=3$
Or,
$y+1=0 \Rightarrow y=-1$
Hence the curve intersects y-axis at $(0,3)$ and $(0,-1)$.
(ii) $2x^{2}+5xy+2y^{2}-8=0$
Sol.
$2x^{2}+5xy+2y^{2}-8=0$ ….. (i)
At x-axis, $y=0$
When $y=0$, (i) $\Rightarrow 2x^{2}-8=0$
$\Rightarrow 2x^{2}=8$
$\Rightarrow x^{2}=4$
$\Rightarrow x=\pm 2$
Hence the curve intersects x-axis at $(2,0)$ and $(-2,0)$.
At y-axis, $x=0$
When $x=0$, (i) $\Rightarrow 2y^{2}-8=0$
$\Rightarrow 2y^{2}=8$
$\Rightarrow y^{2}=4$
$\Rightarrow y=\pm 2$
Hence, the curve intersects y-axis at $(0,2)$ and $(0,-2)$.
(iii) $x^{2}+y^{2}-2x-4y=0$
Sol.
$x^{2}+y^{2}-2x-4y=0$ ….. (i)
At x-axis, $y=0$
When $y=0$, (i) $\Rightarrow x^{2}-2x=0$
$\Rightarrow x(x-2)=0$
$\therefore x=0, 2$
Hence, the curve intersects x-axis at $(0,0)$ and $(2,0)$.
At y-axis, $x=0$
When $x=0$, (i) $\Rightarrow y^{2}-4y=0$
$\Rightarrow y(y-4)=0$
$\therefore y=0, 4$
Hence, the curve intersects y-axis at $(0,0)$ and $(0,4)$.
(iv) $2x^{2}-xy-y^{2}-7x-2y+3=0$
Sol.
$2x^{2}-xy-y^{2}-7x-2y+3=0$ ….. (i)
At x-axis, $y=0$
When $y=0$, (i) $\Rightarrow 2x^{2}-7x+3=0$
$\Rightarrow 2x^{2}-6x-x+3=0$
$\Rightarrow 2x(x-3)-1(x-3)=0$
$\Rightarrow (x-3)(2x-1)=0$
Either,
$x-3=0 \Rightarrow x=3$
Or,
$2x-1=0 \Rightarrow 2x=1 \Rightarrow x=\frac{1}{2}$
Hence the curve intersects x-axis at $(3,0)$ and $(\frac{1}{2},0)$.
At y-axis, $x=0$
When $x=0$, (i) $\Rightarrow -y^{2}-2y+3=0$
$\Rightarrow y^{2}+2y-3=0$
$\Rightarrow y^{2}+3y-y-3=0$
$\Rightarrow y(y+3)-1(y+3)=0$
$\Rightarrow (y+3)(y-1)=0$
Either,
$y+3=0 \Rightarrow y=-3$
Or,
$y-1=0 \Rightarrow y=1$
Hence, the curve intersects y-axis at $(0,-3)$ and $(0,1)$.