Find $25 x^{2}+16 y^{2}$, if $5 x+4 y=8$ and $x y=1$.
Given :
$5x+4y=8$ and $xy=1$.
To find :
We have to find the value of $25 x^{2}+16 y^{2}$.
Solution :
We know that, $(a+b)^2 = a^2+b^2+2ab$.
$(5x+4y)^2=25x^2 + 16y^2+2 (5x)(4y)$
$(5x+4y)^2= 25 x^2 + 16y^2+40xy$.........(i)
Substitute $5x+4y=8$ and $xy=1$ in (i),
$8^2 = 25x^2+16y^2+40(1)$
$64=25x^2+16y^2+40$
$64-40=25x^2+16y^2$
$25x^2+16y^2=24$
Therefore, the value of $25x^2+16y^2$ is 24.
- Related Articles
- Factorise:(i) \( 4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z \)(ii) \( 2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z \)
- If \( 2 x+y=23 \) and \( 4 x-y=19 \), find the values of \( 5 y-2 x \) and \( \frac{y}{x}-2 \).
- \Find $(x +y) \div (x - y)$. if,(i) \( x=\frac{2}{3}, y=\frac{3}{2} \)(ii) \( x=\frac{2}{5}, y=\frac{1}{2} \)(iii) \( x=\frac{5}{4}, y=\frac{-1}{3} \)(iv) \( x=\frac{2}{7}, y=\frac{4}{3} \)(v) \( x=\frac{1}{4}, y=\frac{3}{2} \)
- If $\frac{x+1}{y} = \frac{1}{2}, \frac{x}{y-2} = \frac{1}{2}$, find x and y.
- Simplify: \( \frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+\frac{1}{2} x y \).
- Factorise:\( 4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z \)
- Solve the following pairs of equations by reducing them to a pair of linear equations:(i) \( \frac{1}{2 x}+\frac{1}{3 y}=2 \)\( \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6} \)(ii) \( \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \)\( \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \)(iii) \( \frac{4}{x}+3 y=14 \)\( \frac{3}{x}-4 y=23 \)(iv) \( \frac{5}{x-1}+\frac{1}{y-2}=2 \)\( \frac{6}{x-1}-\frac{3}{y-2}=1 \)(v) \( \frac{7 x-2 y}{x y}=5 \)\( \frac{8 x+7 y}{x y}=15 \),b>(vi) \( 6 x+3 y=6 x y \)\( 2 x+4 y=5 x y \)4(vii) \( \frac{10}{x+y}+\frac{2}{x-y}=4 \)\( \frac{15}{x+y}-\frac{5}{x-y}=-2 \)(viii) \( \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \)\( \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8} \).
- If $x^2 + y^2 = 29$ and $xy = 2$, find the value of(i) $x + y$(ii) $x - y$(iii) $x^4 + y^4$
- If $x = -2$ and $y = 1$, by using an identity find the value of the following:\( \left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right) \)
- Factorize:$4(x - y)^2 - 12(x -y) (x + y) + 9(x + y)^2$
- A pair of linear equations which has a unique solution \( x=2, y=-3 \) is(A) \( x+y=-1 \)\( 2 x-3 y=-5 \)(B) \( 2 x+5 y=-11 \)\( 4 x+10 y=-22 \)(C) \( 2 x-y=1 \)\( 3 x+2 y=0 \)(D) \( x-4 y-14=0 \)\( 5 x-y-13=0 \)
- If $x=1,\ y=2$ and $z=5$, find the value of $x^{2}+y^{2}+z^{2}$.
- If $x = 3$ and $y = -1$, find the values of each of the following using in identity:\( \left(9 y^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right) \)
- Check if the following is true.\( (x+y)^{2}=(x-y)^{2}+4 x y \)
- If $x + y = 4$ and $xy = 2$, find the value of $x^2 + y^2$.
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies