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If x+y=16 and xy=48, find the value of x2+y2.
144
150
164
160
160
Use the identity x2+y2=(x+y)2−2xy. Substitute the given values 162−2(48)=256−96=160.
x + y = 16, xy = 48
(x+y)2=x2+y2+2xy
Use the identity x2+y2=(x+y)2−2xy. Substitute the given values 162−2(48)=256−96=160.
Students often calculate (x+y)2 and forget to subtract 2xy, or incorrectly perform the subtraction of 2xy instead of 2xy.
State the algebraic identity
To find x2+y2, we use the square of the sum formula: (x+y)2=x2+y2+2xy
(x+y)2=x2+y2+2xy
Rearrange to isolate the target expression
Rearrange the identity to solve for x2+y2: x2+y2=(x+y)2−2xy
x2+y2=(x+y)2−2xy
Substitute the given values
Substitute x+y=16 and xy=48 into the formula: x2+y2=(16)2−2(48)
x2+y2=256−96
Final Calculation
Perform the subtraction to get the result: x2+y2=160
x2+y2=160
D is correct because 160 is the result of calculating 162−2(48).
This identity is fundamental for solving quadratic equations and finding the sum of roots or differences of squares in competitive examinations.
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