Sum of the solutions of the equation [x^2] - | vedclass

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

Question

(C) The given equation is $[x^2] = 2x - 1$.Since $[x^2]$ is an integer, $2x - 1$ must be an integer, which implies $2x$ is an integer.By definition of

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The given equation is $[x^2] = 2x - 1$.Since $[x^2]$ is an integer, $2x - 1$ must be an integer, which implies $2x$ is an integer.By definition of the greatest integer function, $x^2 - 1 Substituting $[x^2] = 2x - 1$, we get $x^2 - 1 From $x^2 - 1 From $2x - 1 \le x^2$, we get $x^2 - 2x + 1 \ge 0$, which means $(x - 1)^2 \ge 0$, which is true for all real $x$.Case $1$: $0 Case $2$: $1 \le x Case $3$: $\sqrt{2} \le x Case $4$: $\sqrt{3} \le x The solutions are $1/2, 1, 3/2$. The sum is $1/2 + 1 + 3/2 = 3$.

Sum of the solutions of the equation $[x^2] - 2x + 1 = 0$ is (where $[.]$ denotes the greatest integer function).

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + ax + b = 0$,then the value of $\alpha^3 + \beta^3$ is equal to:

Solve the given two equations and provide the correct answer from the given options.
$I.$ $391 x^{2} + 1344 x + 1073 = 0$
$II.$ $437 y^{2} + 1074 y + 589 = 0$

Let $a, b, c \in R$ and $\alpha, \beta$ be the real roots of the equation $ax^2 + bx + c = 0$. If $a < 0, b > 0, c > 0$ and $\alpha < \beta$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} - 468 = 1729$
$II.$ $y^{2} - 1733 + 1564 = 0$

If the ratio of the roots of $ax^2 + 2bx + c = 0$ is the same as the ratio of the roots of $px^2 + 2qx + r = 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module