Question

the sum of squares of two consecutive multiples of 7 is 163 find the multiples.​

Answer 1

Step-by-step explanation:

Let the two multiples of 7 be k and (k+7).

It is given that the sum of squares of k and (k+7) is 163

${k}^{2} + {(k + 7)}^{2} = 163 = > {k}^{2} + {k}^{2} +$

$14k + 49 = 163 = > 2 {k}^{2} + 14k +$

$49 - 163 = > {k}^{2} + 7k - 57 = 0$

Roots are ( -7+√49+228)/2 and (-7-√49+228)/2. Roots are irrational so it is not possible. Hope it helps you.