Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
If x+y=20 and xy=84, find the value of x2+y2.
264
212
232
244
232
Use the identity x2+y2=(x+y)2−2xy. Square 20 to get 400, then subtract twice the product (2 * 84 = 168). 400 - 168 = 232.
x + y = 20, xy = 84
(x+y)2=x2+y2+2xy
Use the identity x2+y2=(x+y)2−2xy. Square 20 to get 400, then subtract twice the product (2 * 84 = 168). 400 - 168 = 232.
Students often forget the factor of 2 in the 2xy term or calculate (x+y)2−xy instead of (x+y)2−2xy.
State the algebraic identity
To find x2+y2, we use the algebraic expansion of the square of a binomial: (x+y)2=x2+2xy+y2.
(x+y)2=x2+y2+2xy
Rearrange for the target expression
Isolate x2+y2 by subtracting 2xy from both sides of the equation.
x2+y2=(x+y)2−2xy
Substitute the known values
Substitute the given values x+y=20 and xy=84 into the rearranged identity.
x2+y2=(20)2−2(84)
Calculate the final result
Perform the arithmetic: 202=400 and 2×84=168. Subtracting these yields 400−168=232.
x2+y2=400−168=232
C is correct because the identity x2+y2=(x+y)2−2xy results in 400−168=232.
This identity is fundamental for solving quadratic equations and finding the roots of polynomials. It frequently appears in coordinate geometry when finding the squared distance from the origin (x2+y2).
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download • 100K+ Aspirants