The sum of the squares of two natural consecu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the two natural consecutive odd numbers be $n$ and $(n+2)$.According to the question,$n^{2} + (n+2)^{2} = 394$$n^{2} + n^{2} + 4 + 4n = 394$$2

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) Let the two natural consecutive odd numbers be $n$ and $(n+2)$.According to the question,$n^{2} + (n+2)^{2} = 394$$n^{2} + n^{2} + 4 + 4n = 394$$2n^{2} + 4n - 390 = 0$Dividing by $2$:$n^{2} + 2n - 195 = 0$Factoring the quadratic equation:$n^{2} + 15n - 13n - 195 = 0$$n(n+15) - 13(n+15) = 0$$(n-13)(n+15) = 0$Since $n$ must be a natural number,$n = 13$.The two numbers are $13$ and $15$.The sum of the numbers is $13 + 15 = 28$.

The sum of the squares of two natural consecutive odd numbers is $394$. The sum of the numbers is

Similar Questions

If $a, b, c \in \mathbb{Q}$,then the roots of the equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$ are

Solve the given two equations and select the correct answer from the given options.
$I.$ $8x^2 + 6x = 5$
$II.$ $12y^2 - 22y + 8 = 0$

Which of the following equations has equal roots?

If $(x^{3}-y^{3}):(x^{2}+xy+y^{2})=5:1$ and $(x^{2}-y^{2}):(x-y)=7:1$,then the ratio $2x:3y$ equals

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$

Vedclass Test Series

Exam Paper Generator

Online Exam Module