Respuesta :
Answer:
[tex]\text{Given:}\\y\text{ is the mean proportional between }x\text{ and }z.\\\text{i.e. }y^2=xz\\\text{We have to prove:}\\\text{}xy+yz\text{ is the mean proportional between }x^2+y^2\text{ and }y^2+z^2.\\\text{i.e. }(xy+yz)^2=(x^2+y^2)(y^2+z^2)[/tex]
[tex]\text{Now,}\\\text{L.H.S. = }(xy+yz)^2=y^2(x+z)^2=xz(x+z)^2\\\text{R.H.S. = }(x^2+y^2)(y^2+z^2)=(x^2+xz)(xz+z^2)=x(x+z)z(x+z)=xz(x+z)^2[/tex]
[tex]\therefore\ \text{L.H.S. = R.H.S. proved}[/tex]
Alternative method
[tex]\text{Solution:}\\\\\text{Let }\dfrac{x}{y}=\dfrac{y}{z}=k\\\\\text{or, }x=yk\text{ and }y=zk\\\text{or, }x=(zk)k\text{ and }y=zk\\\therefore\ x=zk^2\text{ and }y=zk[/tex]
[tex]\text{We have to prove:}\\(xy+yz)^2=(x^2+y^2)(y^2+z^2)\\\\\text{L.H.S.}=(xy+yz)^2=(zk^2.zk+zk.z)^2=(z^2k^3+z^2k)^2=\{z^2k(k^2+1)\}^2\\\\\text{}\qquad\quad =z^4k^2(k^2+1)^2[/tex]
[tex]\text{R.H.S. = }(x^2+y^2)(y^2+z^2)\\\\\text{}\qquad\quad=\{(zk^2)^2+(zk)^2\}\{(zk)^2+z^2\}\\\\\text{}\qquad\quad=(z^2k^4+z^2k^2)(z^2k^2+z^2)\\\\\text{}\qquad\quad=z^2k^2(k^2+1)z^2(k^2+1)\\\\\text{}\qquad\quad=z^4k^2(k^2+1)^2[/tex]
Final answer:
The proof establishes that if y is the mean proportional between x and z, then xy · yz acts as the mean proportional between x^2y^2 and y^2z^2, validated through substitution and algebraic manipulation.
Explanation:
If y is the mean proportional between x and z, this means that y^2 = xz. The question then evolves into proving that xy · yz is the mean proportional between x^2y^2 and y^2z^2. This proposition relies on understanding the properties of proportions and manipulations of algebraic expressions.
First, remember that if a is the mean proportional between b and c, then a^2 = bc. Given that y is the mean proportional between x and z, we have y^2 = xz. We need to verify whether xy · yz behaves similarly in relation to x^2y^2 and y^2z^2.
- Substitute y^2 = xz into xy · yz, we get xyz^2. Now, replacing y^2 with xz, we have x^2z^2, which is the product of xz by itself.
- For the quantities x^2y^2 and y^2z^2, using the relation y^2 = xz, these can be rewritten as x^2(xz) and (xz)z^2, which simplifies to x^3z and xz^3 respectively.
- Looking at xy · yz = x^2z^2, it indeed serves as the mean proportional between x^3z and xz^3 because (x^2z^2)^2 = x^4z^4 = (x^3z)(xz^3).
Hence, it's shown that if y is the mean proportion between x and z, then xy · yz indeed serves as the mean proportion between x^2y^2 and y^2z^2.
