Consider the equation x^2+x-n = 0,where n in | vedclass

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

Question

(A) The given equation is $x^2+x-n=0$. For the roots to be integers,the discriminant $D = b^2-4ac = 1^2 - 4(1)(-n) = 1+4n$ must be a perfect square of

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The given equation is $x^2+x-n=0$. For the roots to be integers,the discriminant $D = b^2-4ac = 1^2 - 4(1)(-n) = 1+4n$ must be a perfect square of an odd integer. Let $1+4n = (2k+1)^2$ for some integer $k$. $1+4n = 4k^2 + 4k + 1 $ $\Rightarrow 4n = 4k(k+1) $ $\Rightarrow n = k(k+1)$. Given $n \in [5, 100]$,we have $5 \le k(k+1) \le 100$. For $k=2, n=2(3)=6$. For $k=3, n=3(4)=12$. For $k=4, n=4(5)=20$. For $k=5, n=5(6)=30$. For $k=6, n=6(7)=42$. For $k=7, n=7(8)=56$. For $k=8, n=8(9)=72$. For $k=9, n=9(10)=90$. For $k=10, n=10(11)=110 > 100$. Thus,the possible values of $n$ are $6, 12, 20, 30, 42, 56, 72, 90$. The total number of such values is $8$.

Consider the equation $x^2+x-n = 0$,where $n \in N$ and $n \in [5, 100]$. The total number of different values of $n$ such that the given equation has integral roots is:

Similar Questions

What is the number of real solutions for the equation $x^2 - 3|x| + 2 = 0$?

Let $f(x)=ax^{2}+bx+c$ and $g(x)=px^{2}+qx+r$ such that $f(1)=g(1)$, $f(2)=g(2)$ and $f(3)-g(3)=2$. Then, $f(4)-g(4)$ is:

The solution of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$ is

Solve the equation $x^{2}+3x+9=0$.

If $f(x)$ is a second degree polynomial such that $f(x) \geq 0$ for all $x \in R$,$f(-3) = 0$,and $f(0) = 18$,then find $f(3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module