Question

The sum of the squares of two numbers is 1088. If one of the numbers is 8, find the other number.

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Answer 1

Answer:

et no be x

8²+x²=1088

x²=1088-64

x²= 1024

x=√1024

x= 32

Step-by-step explanation:

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Answer 2

Given :

The sum of the squares of two numbers is 1088.

One of the number is 8 .

To Find :

Other number .

Solution :

$\longmapsto\tt{Let\:other\:number\:be={(b)}^{2}}$

A.T.Q :

$\longmapsto\tt{{(8)}^{2}+{(b)}^{2}=1088}$

$\longmapsto\tt{64+{(b)}^{2}=1088}$

$\longmapsto\tt{{b}^{2}=1088-64}$

$\longmapsto\tt{{b}^{2}=1024}$

$\longmapsto\tt{b=\sqrt{1024}}$

$\longmapsto\tt\bf{b=\pm{32}}$

So , The other number is ±32 .

VERIFICATION :

$\longmapsto\tt{{(8)}^{2}+{(b)}^{2}=1088}$

$\longmapsto\tt{{(8)}^{2}+{(32)}^{2}=1088}$

$\longmapsto\tt{64+1024=1088}$

$\longmapsto\tt\bf{1088=1088}$

HENCE VERIFIED

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