If the roots of the equation 8x^3 - 14x^2 + 7 | vedclass

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

Question

(A) Let the roots of the cubic equation be $\frac{a}{r}, a, ar$. According to the properties of roots of a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$,t

Vedclass · Accepted Answer

$1, \frac{1}{2}, \frac{1}{4}$

Vedclass · Answer

(A) Let the roots of the cubic equation be $\frac{a}{r}, a, ar$. According to the properties of roots of a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$,the product of the roots is given by $-\frac{D}{A}$. Here,$A = 8, B = -14, C = 7, D = -1$. Product of roots: $(\frac{a}{r}) 	imes a 	imes (ar) = a^3 = -(\frac{-1}{8}) = \frac{1}{8}$. Thus,$a^3 = \frac{1}{8} \implies a = \frac{1}{2}$. Since $a = \frac{1}{2}$ is a root,it must satisfy the equation: $8(\frac{1}{2})^3 - 14(\frac{1}{2})^2 + 7(\frac{1}{2}) - 1 = 8(\frac{1}{8}) - 14(\frac{1}{4}) + \frac{7}{2} - 1 = 1 - 3.5 + 3.5 - 1 = 0$. Dividing the polynomial by $(x - \frac{1}{2})$ or $(2x - 1)$,we get $4x^2 - 5x + 1 = 0$. Factoring $4x^2 - 5x + 1 = (4x - 1)(x - 1) = 0$,we get $x = 1$ and $x = \frac{1}{4}$. The roots are $1, \frac{1}{2}, \frac{1}{4}$,which are in $G.P.$ with common ratio $r = \frac{1}{2}$.

If the roots of the equation $8x^3 - 14x^2 + 7x - 1 = 0$ are in $G.P.$,then the roots are

Similar Questions

Find the $12^{\text{th}}$ term of a $G.P.$ whose $8^{\text{th}}$ term is $192$ and the common ratio is $2$.

Find the sum of the first $n$ terms and the sum of the first $5$ terms of the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$

The sum of a few terms of a geometric series is $728$. If the common ratio is $3$ and the last term is $486$,then the first term of the series will be:

For a $G$.$P$.,if $S_n = \frac{4^n - 3^n}{3^n}$,then $t_2 = ........$

Find the fractional value of $0.125125125 \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module