Suppose a, b denote the distinct real roots o | vedclass

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

Question

(D) Given $x^2+20x-2020=0$ has roots $a, b$. By Vieta's formulas,$a+b = -20$ and $ab = -2020$.Given $x^2-20x+2020=0$ has roots $c, d$. By Vieta's form

Vedclass · Accepted Answer

$16000$

Vedclass · Answer

(D) Given $x^2+20x-2020=0$ has roots $a, b$. By Vieta's formulas,$a+b = -20$ and $ab = -2020$.Given $x^2-20x+2020=0$ has roots $c, d$. By Vieta's formulas,$c+d = 20$ and $cd = 2020$.We need to evaluate $E = ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$.Expanding the expression: $E = a^2c - ac^2 + a^2d - ad^2 + b^2c - bc^2 + b^2d - bd^2$.Grouping terms: $E = a^2(c+d) + b^2(c+d) - c^2(a+b) - d^2(a+b)$.$E = (c+d)(a^2+b^2) - (a+b)(c^2+d^2)$.Using $a^2+b^2 = (a+b)^2 - 2ab$ and $c^2+d^2 = (c+d)^2 - 2cd$:$a^2+b^2 = (-20)^2 - 2(-2020) = 400 + 4040 = 4440$.$c^2+d^2 = (20)^2 - 2(2020) = 400 - 4040 = -3640$.Substituting these values: $E = (20)(4440) - (-20)(-3640)$.$E = 88800 - 72800 = 16000$.

Suppose $a, b$ denote the distinct real roots of the quadratic polynomial $x^2+20x-2020$ and suppose $c, d$ denote the distinct complex roots of the quadratic polynomial $x^2-20x+2020$. Then the value of $ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$ is

Similar Questions

The probability of selecting integers $a \in [-5, 30]$ such that $x^{2}+2(a+4)x-5a+64 > 0$ for all $x \in \mathbb{R}$ is:

Given the equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$ has roots $\sin \theta$ and $\cos \theta$ $(0 < \theta < \frac{\pi}{2})$,then the maximum value of $(a + \sin \theta)$ is-

If $x$ is real,then the difference between the greatest and least values of $\frac{x^2-x+1}{x^2+x+1}$ is

Both equations $x^2 + b^2 = 1 - 2bx$ and $x^2 + a^2 = 1 - 2ax$ have exactly one root each,and they share the same root. Then:

Let $r$ be a root of the equation $x^2+2x+6=0$. The value of $(r+2)(r+3)(r+4)(r+5)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module