If the product of the roots of the equation { | vedclass

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

Question

(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.Using the property of logarithms,${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$.Thus,the

Vedclass · Accepted Answer

$k = 2$

Vedclass · Answer

(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.Using the property of logarithms,${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$.Thus,the equation becomes ${x^2} - 3kx + 2{k^2} - 1 = 0$.Since the product of the roots is $7$,we have $2{k^2} - 1 = 7$,which implies $2{k^2} = 8$,so ${k^2} = 4$,giving $k = \pm 2$.Since the term $\log k$ is present in the original equation,$k$ must be positive,so $k = 2$.For the roots to be real,the discriminant $D = b^2 - 4ac$ must be $\ge 0$.Here,$D = (-3k)^2 - 4(1)(2{k^2} - 1) = 9{k^2} - 8{k^2} + 4 = {k^2} + 4$.Since ${k^2} + 4$ is always greater than $0$ for any real $k$,the roots are real for $k = 2$.

If the product of the roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then for what value of $k$ will its roots be real?

Similar Questions

If $4a - \frac{4}{a} + 3 = 0$,then the value of $a^{3} - \frac{1}{a^{3}} + 3$ is equal to?

If one root of the equation $ax^2 + bx + c = 0$ is the square of the other,then $a(c - b)^3 = cX$,where $X$ is

If the roots of the equation $ax^2 + bx + c = 0$ are $l$ and $2l$,then:

The solution of $2-x=\frac{x-2}{x}$ would include

Solve the given two equations and select the correct answer from the given options.
$I.$ $(441)^{\frac{1}{2}} x^{2} - 111 = (15)^{2}$
$II.$ $\sqrt{121} y^{2} + (6)^{3} = 260$

Vedclass Test Series

Exam Paper Generator

Online Exam Module