Question

Sum of squares of two consecutive positive even integers is 100. Find those numbers by using quadratic equations.

Answer 1

Hint: For this type of problem to solve we have to take two consecutive numbers, square them and add them and by solving get the value of x. After that to find the other number add one to it which gives us the final answer.

Complete step-by-step answer:
Considering two consecutive numbers as \[x,x+1\].
By squaring and adding those two we get
\[{{x}^{2}}+{{(x+1)}^{2}}=100\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
By simplifying the equation by using the formulas we get,
\[{{x}^{2}}+{{x}^{2}}+2x+1=100\]
Writing the equation in terms of quadratic equation we get
\[2{{x}^{2}}+2x-99=0\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
To solve the quadratic equation we use the formula
\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. . . . . . . . . . . . . . . . . . . . . (a)
By substituting the values in (a) we get, a=2, b=2, c= -99,
\[\dfrac{-2\pm \sqrt{{{2}^{2}}-4(2)\left( -99 \right)}}{2(2)}\]
\[\dfrac{-2\pm \sqrt{796}}{4}\]
\[\dfrac{-2\pm 28.21}{4}\]
\[{{x}_{1}}=-7.553\]
\[{{x}_{2}}=6.5533\]
Therefore the value of x is \[6.5533\]
The value of x+1 is \[7.553\]

Note: Above we have got two roots but took only the positive because it was mentioned in the question that given two positive integers. So we have taken positive values of the two roots. The consecutive number is formed by adding 1 to x.