Which of the following conditions implies tha | vedclass

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

Question

(C) Given the quadratic equation $\frac{1}{4}x^2 + bx + c = 0$. Multiplying the entire equation by $4$,we get $x^2 + 4bx + 4c = 0$. Using the quadrati

Vedclass · Accepted Answer

$b^2 - c$ is the square of an integer and $b$ is an integer

Vedclass · Answer

(C) Given the quadratic equation $\frac{1}{4}x^2 + bx + c = 0$. Multiplying the entire equation by $4$,we get $x^2 + 4bx + 4c = 0$. Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ for the equation $x^2 + 4bx + 4c = 0$,where $A=1, B=4b, C=4c$: $x = \frac{-4b \pm \sqrt{(4b)^2 - 4(1)(4c)}}{2(1)}$ $x = \frac{-4b \pm \sqrt{16b^2 - 16c}}{2}$ $x = \frac{-4b \pm 4\sqrt{b^2 - c}}{2}$ $x = -2b \pm 2\sqrt{b^2 - c}$. For $x$ to be an integer,$b$ must be an integer and $b^2 - c$ must be a perfect square of an integer.

Which of the following conditions implies that the roots of the equation $\frac{1}{4}x^2 + bx + c = 0$ are integers?

Similar Questions

The $x$-coordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3|x| + 2 = 0$ and the $y$-coordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$. Then the possible vertices of the square are:

The product of all the real roots of $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ is

Find the number of real solutions for the equation $\left( \frac{5}{7} \right)^x = -x^2 + 2x - 3$.

If the roots of the equation $x^2 + px + q = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,then the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ will be

Let $f(x) = x^3 + a x^2 + b x + c$ be a polynomial with integer coefficients. If the roots of $f(x)$ are integers and are in Arithmetic Progression,then '$a$' cannot take the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module