Let [x] be the greatest integer less than or | vedclass

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

Question

(C) Given the equation $[x^2] = x + 1$. Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer. Let $x = n$,where $n \

Vedclass · Accepted Answer

no solution

Vedclass · Answer

(C) Given the equation $[x^2] = x + 1$. Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer. Let $x = n$,where $n \in \mathbb{Z}$. The equation becomes $[n^2] = n + 1$. Since $n^2$ is an integer,$[n^2] = n^2$. So,$n^2 = n + 1$,which implies $n^2 - n - 1 = 0$. The roots of this quadratic equation are $n = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $\frac{1 \pm \sqrt{5}}{2}$ are not integers,there is no integer $n$ that satisfies the equation. Therefore,the equation $[x^2] = x + 1$ has no solution.

Let $[x]$ be the greatest integer less than or equal to $x$ for a real number $x$. Then the equation $[x^2] = x + 1$ has:

Similar Questions

Let $\alpha_1, \alpha_2, \ldots, \alpha_7$ be the roots of the equation $x^7+3x^5-13x^3-15x=0$ and $|\alpha_1| \geq |\alpha_2| \geq \ldots \geq |\alpha_7|$. Then $\alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6$ is equal to $..................$.

The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$ is equal to:

The value of $m$ for which the equation $\frac{a}{x + a + m} + \frac{b}{x + b + m} = 1$ has roots equal in magnitude but opposite in sign is

Solve the equation $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$.

If $\alpha, \beta$ are the irrational roots of the equation $3p^2x^3 + px^2 + qx + 3 = 0$ when $p = 1$ and $q = -7$,then $|\alpha - \beta| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module