If a, b, c, d and p are distinct real numbers | vedclass

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

Question

(D) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$.This can be rearranged as:$(a^2 p^2 - 2abp + b^2) + (b

Vedclass · Accepted Answer

$a, b, c, d$ are in $G.P.$

Vedclass · Answer

(D) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$.This can be rearranged as:$(a^2 p^2 - 2abp + b^2) + (b^2 p^2 - 2bpc + c^2) + (c^2 p^2 - 2cpd + d^2) \le 0$.This simplifies to:$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$.Since the sum of squares of real numbers can only be non-negative,the only way for the sum to be less than or equal to $0$ is if each term is exactly $0$.Therefore,$ap - b = 0$,$bp - c = 0$,and $cp - d = 0$.This implies $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.This is the definition of a Geometric Progression $(G.P.)$,where $a, b, c, d$ are in $G.P.$ with common ratio $p$.

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$,then

Similar Questions

If the roots of the equation $Ax^2 + Bx + C = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2, \beta^2$,then the value of $p$ will be

Solve the given two equations and select the correct option.
$I.$ $2x^2 + 3x + 1 = 0$
$II.$ $12y^2 + 7y + 1 = 0$

If the roots of the equation $ax^2 + bx + c = 0$ are real and of the form $\frac{\alpha}{\alpha - 1}$ and $\frac{\alpha + 1}{\alpha}$,then the value of $(a + b + c)^2$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x + 3y = 77$
$II.$ $2x + 5y = (2601)^{1/2}$

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module