If the set of all a in R,for which the equati | vedclass

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

Question

(C) The given equation is $2x^2 + (a-5)x + (15-3a) = 0$. For the equation to have no real roots,the discriminant $D < 0$. $D = (a-5)^2 - 4(2)(15-3a) <

Vedclass · Accepted Answer

$2139$

Vedclass · Answer

(C) The given equation is $2x^2 + (a-5)x + (15-3a) = 0$. For the equation to have no real roots,the discriminant $D $D = (a-5)^2 - 4(2)(15-3a) $a^2 - 10a + 25 - 8(15-3a) $a^2 - 10a + 25 - 120 + 24a $a^2 + 14a - 95 $(a+19)(a-5) Thus,$a \in (-19, 5)$,so $\alpha = -19$ and $\beta = 5$. The set $X = \{x \in Z : -19 We need to calculate $\sum_{x \in X} x^2 = \sum_{x=-18}^{4} x^2$. This is equal to $(1^2 + 2^2 + 3^2 + 4^2) + 0^2 + (1^2 + 2^2 + \ldots + 18^2)$. Sum of squares formula: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. Sum $= \frac{4(5)(9)}{6} + \frac{18(19)(37)}{6} = 30 + 2109 = 2139$.

If the set of all $a \in R$,for which the equation $2x^2 + (a-5)x + (15-3a) = 0$ has no real root,is the interval $(\alpha, \beta)$,and $X = \{x \in Z : \alpha < x < \beta\}$,then $\sum_{x \in X} x^2$ is equal to

Similar Questions

If $1-i$ is a root of the equation $x^2+ax+b=0$ where $a$ and $b$ are real numbers,then $b$ is equal to

The roots of the given equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ are

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

The sum of the real roots of the equation $x^4-2x^3+x-380=0$ is

The solution of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module