Question

the sum of squares of three consecutive natural numbers is 434. take the middle number as x and frame an equation in x, and solve it to find the three numbers

Answer 1

Answer:

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Answer 2

Given:

The Sum of three consecutive natural numbers is 434.

To Find:

Frame an equation to find three numbers.

Solution:

Suppose the three consecutive natural numbers are $x-1,x$ and$x+1.$

here, the sum of their squares = $(x-1)^{2}+x^{2} +(x+1)^{2}$= 434

now,

$(x^{2} -2x+1)+x^{2} +(x^{2} +2x+1)=434\\3x^{2}+2-434=0\\ 3x^{2}-432=0 \\3(x^{2} -144)=0\\3(x+12)(x-12)=0\\x=12$

The value of x=12, now we will put the value x in the other two equations.

Hence, the natural numbers are 11, 12, and 13.