Let S be the set of all possible integral val | vedclass

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

Question

(A) For the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ to be rational,the discriminant $D$ must be a perfect square of a rational num

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ to be rational,the discriminant $D$ must be a perfect square of a rational number. Since $\lambda$ is an integer,$D$ must be a perfect square of an integer.$D = b^2 - 4ac = (13)^2 - 4(\lambda)(7) = 169 - 28\lambda$.Given $\lambda \in (-3, 7)$,the possible integer values for $\lambda$ are $\{-2, -1, 0, 1, 2, 3, 4, 5, 6\}$.We test these values:If $\lambda = -2$,$D = 169 - 28(-2) = 169 + 56 = 225 = (15)^2$ (Perfect square).If $\lambda = -1$,$D = 169 - 28(-1) = 197$ (Not a perfect square).If $\lambda = 0$,the equation becomes $13x + 7 = 0$,which gives $x = -7/13$ (Rational).If $\lambda = 1$,$D = 169 - 28 = 141$ (Not a perfect square).If $\lambda = 2$,$D = 169 - 56 = 113$ (Not a perfect square).If $\lambda = 3$,$D = 169 - 84 = 85$ (Not a perfect square).If $\lambda = 4$,$D = 169 - 112 = 57$ (Not a perfect square).If $\lambda = 5$,$D = 169 - 140 = 29$ (Not a perfect square).If $\lambda = 6$,$D = 169 - 168 = 1 = (1)^2$ (Perfect square).Thus,the set $S = \{-2, 0, 6\}$.The sum of the elements in $S$ is $-2 + 0 + 6 = 4$.

Let $S$ be the set of all possible integral values of $\lambda$ in the interval $(-3, 7)$ for which the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ are all rational numbers. Then the sum of the elements in $S$ is

Similar Questions

The quadratic equation $x^2+15|x|+14=0$ has

Given $f(x) = ax^2 + bx + c$,$g(x) = a_1x^2 + b_1x + c_1$,and $p(x) = f(x) - g(x)$. If $p(x) = 0$ only for $x = -1$ and $p(-2) = 2$,what is the value of $p(2)$? Assume $a \neq a_1 \neq 0$.

Find the product of all real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$.

The value of $m$ for which the equation $\frac{a}{x + a + m} + \frac{b}{x + b + m} = 1$ has roots equal in magnitude but opposite in sign is

The product of real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module