Exercise 4.3 | Class 10 | Advanced Mathematics | Quadratic Equation

Solve the following equations:
(1) $~~ 3x-2y=2$, $9x^2+4y^2=2$
Solution:
Let, $3x-2y=2 ⇒ x = \frac{2+2y}{3}$ ...(1)
and $9x^2+4y^2=2$ ...(2)
⇒ $9[\frac{2+2y}{3}]^2+4y^2 = 2$   [using (1)]
⇒ $9[\frac{4+4y^2+8y}{9}]+4y^2 = 2$
⇒ $4+4y^2+8y+4y^2=2$
⇒ $8y^2+8y+2 = 0$
⇒ $4y^2+4y+1 = 0$
⇒ $(2y+1)^2 = 0$
⇒ $y = -\frac{1}{2}$ or $-\frac{1}{2}$
When $y = -\frac{1}{2}$
(1)⇒ $x = \frac{2+2(-\frac{1}{2})}{3} = \frac{2-1}{3} = \frac{1}{3}$
Since both the value of $y$ are equal, the required solutions are $(\frac{1}{3}, -\frac{1}{2})$ and $(\frac{1}{3}, -\frac{1}{2})$



(8) $~~ x+y+\sqrt{xy}=28$, $x^2+y^2+xy=336$
Solution:
Let, $x+y+\sqrt{xy}=28$ ...(1)
and $x^2+y^2+xy=336$ ...(2)
⇒ $(x+y)^2-2xy+xy = 336$
⇒ $(x+y)^2-xy = 336$
⇒ $(x+y)^2-(\sqrt{xy})^2 = 336$
⇒ $(x+y+\sqrt{xy})(x+y-\sqrt{xy}) = 336$
⇒ $28(x+y-\sqrt{xy}) = 336$   [using (1)]
⇒ $x+y - \sqrt{xy}=12$ ...(3)
(1)+(3) ⇒ $x+y+\sqrt{xy}+x+y-\sqrt{xy} = 28+12$
⇒ $2(x+y) = 40$
⇒ $x+y = 20$
⇒ $x = 20-y$ ...(4)
(2) ⇒ $(20-y)^2+y^2+(20-y) = 336$   [using (4)]
⇒ $400-40y+y^2+y^2-y^2+20y = 336$
⇒ $y^2-16y-4y+64 = 0$
⇒ $y(y-16)-4(y-16) = 0$
⇒ $y = 4$ or $16$
When $y = 4$, (4)⇒ $x = 20-4 = 16$
When $y = 16$, (4)⇒ $x = 20-16 = 4$
⸫ The required solutions are $(4, 16)$ and $(16, 4)$



(10) $~~ \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}} = \frac{10}{3}$, $x+y=10$
Solution:
Let, $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}} = \frac{10}{3}$ ...(1)
and $x+y=10$ ...(2)
⇒ $x = 10-y$ ...(3)

(1)⇒ $\frac{x+y}{\sqrt{xy}} = \frac{10}{3}$
⇒ $\frac{10}{\sqrt{xy}} = \frac{10}{3}$   [using (2)]
⇒ $\sqrt{xy} = 3$
⇒ $\sqrt{(10-y)y} = 3$
⇒ $(\sqrt{10y - y^2})^2 = 3^3$
⇒ $10y - y^2 = 9$
⇒ $y^2-10y+9 = 0$
⇒ $y^2-9y-y+9 = 0$
⇒ $y(y-9)-1(y-9) = 0$
⇒ $(y-1)(y-9) = 0$
⇒ $y = 1$ or $9$
When $y = 1$, (3)⇒ $x = 10-1 = 9$
When $y = 9$, (3)⇒ $x = 10-9 = 1$
⸫ The required solutions are $(4, 16)$ and $(16, 4)$



(12) $~~ x+y = p+q$, $\frac{p}{q}+\frac{q}{y} = 2$
Solution:
Let, $x+y=p+q$ ...(1)
⇒ $x-p = q-y$
⇒ $p-x = y-q$ ...(2)
and $\frac{p}{q}+\frac{q}{y} = 2$
⇒ $\frac{p}{q}-1+\frac{q}{y}-1 = 0$ ...(3)
⇒ $\frac{p-x}{x}+\frac{q-y}{y} = 0$
⇒ $\frac{y-q}{x}+\frac{y-q}{y} = 0$   [using (2)]
⇒ $(y-q)(\frac{1}{x}-\frac{1}{y}) = 0$
Either $y = q$
or $\frac{1}{x}-\frac{1}{y}$
⇒ $\frac{1}{x} = \frac{1}{y}$
⇒ $x = y$
When $y = q$ , (1)⇒ $x+q = p+q ⇒ x = p$
When $y = x$ , (1)⇒ $x+x = p+q ⇒ x = \frac{p+q}{2}$ $$\ ⸫~~ y = \frac{p+q}{2} ~~[⸪~~ x = y]$$ ⸫ The required solutions are $(p, q)$ and $(\frac{p+q}{2}, \frac{p+q}{2})$


(14) $~~ \frac{a}{x}+\frac{b}{y} = 2$, $\frac{a^2}{x^2}+\frac{b^2}{y^2} = 2$
Solution:
Let, $\frac{a}{x}+\frac{b}{y} = 2$ ...(1)
and $\frac{a^2}{x^2}+\frac{b^2}{y^2} = 2$
⇒ $(\frac{a}{x}+\frac{b}{y})^2 -\frac{2ab}{xy} = 2$
⇒ $(2)^2 -\frac{2ab}{xy} = 2$   [using (1)]
⇒ $2 = \frac{2ab}{xy}$
⇒ $\frac{ab}{xy} = 1$
⇒ $\frac{b}{y} = \frac{x}{a}$ ...(2)
(1)⇒ $\frac{a}{x}+\frac{x}{a} = 2$   [using (2)]
⇒ $\frac{a^2+x^2}{ax} = 2$
⇒ $a^2+x^2 = 2ax$
⇒ $a^2+x^2-2ax = 0$
⇒ $(x-a)^2 = 0$
⇒ $(x-a)(x-a) = 0$
⇒ $x = a$ or $a$
When $x = a$ , (2)⇒ $\frac{b}{y}= \frac{a}{a} ⇒ \frac{b}{y}=1 ⇒ y = b$
Since both the value of $x$ are equal, the required solutions are $(a, b)$ and $(a, b)$



(16) $~~ 8.2^{xy}=4^y$, $27.9^x.3^{xy} = 1$
Solution:
Let, $8.2^{xy}=4^y$
⇒ $2^3.2^{xy} = (2^2)^y$
⇒ $2^{3+xy} = 2^{2y}$
⇒ $3+xy = 2y$
⇒ $xy = 2y-3$ ...(1)
and $27.9^x.3^{xy} = 1$
⇒ $3^3.(3^{2x}.3^{xy} = 1$
⇒ $3^{3+2x+xy} = 3^0$
⇒ $3+2x+xy = 0$
⇒ $3+2x+2y-3 = 0$   [using (1)]
⇒ $2(x+y) = 0$
⇒ $x = -y$ ...(2)
(1)⇒ $(-y) y = 2y-3$
⇒ $-y^2-2y+3 = 0$
⇒ $y^2+2y-3 = 0$
⇒ $y^2+3y-y-3 = 0$
⇒ $y(y+3)-1(y+3) = 0$
⇒ $(y+3)(y-1) = 0$
⇒ $y = -3$ or $1$
When $y = -3$ , (2)⇒ $x = -(-3) ⇒ x = 3$
When $y = 1$ , (2)⇒ $x = -(1) ⇒ x = -1$
⸫ The required solutions are $(-1, 3)$ and $1, -3)$




20. Find the coordinates of the points at which x-axis and y-axis intersect the curves represented by the following equations:
(i) $2x^2-3xy+y^2+x-2y-3=0$
Solution:
$2x^2-3xy+y^2+x-2y-3=0$ ...(1)
At x-axis, $y=0$, then
(1) ⇒ $2x^2+x-3=0$
⇒ $2x^2+3x-2x-3=0$
⇒ $x(2x+3)-1(2x+3)=0$
⇒ $(2x+3)(x-1)=0$
⇒ $x=-\frac{3}{2}$ or $x=1$
⸫ The curve intersects x-axis at ($-\frac{3}{2}, 0$) and ($1,0$)

At y-axis, $x=0$, then
(1) ⇒ $y^2-2y-3=0$
⇒ $y^2-3y+y-3=0$
⇒ $y(y-3)+1(y-3)=0$
⇒ $(y-3)(y+1)=0$
⇒ $x=3$ or $x=-1$
⸫ The curve intersects y-axis at ($0, 3$) and ($0, -1$)

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