Let a and b be roots of x^2 - 3x + p = 0 and | vedclass

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

Question

(C) Given that $a, b$ are roots of $x^2 - 3x + p = 0$,we have $a + b = 3$ and $ab = p$.Given that $c, d$ are roots of $x^2 - 12x + q = 0$,we have $c +

Vedclass · Accepted Answer

$17 : 15$

Vedclass · Answer

(C) Given that $a, b$ are roots of $x^2 - 3x + p = 0$,we have $a + b = 3$ and $ab = p$.Given that $c, d$ are roots of $x^2 - 12x + q = 0$,we have $c + d = 12$ and $cd = q$.Since $a, b, c, d$ are in an increasing $G$.$P$.,let the terms be $a, ar, ar^2, ar^3$.Then $a + b = a(1 + r) = 3$ and $c + d = ar^2(1 + r) = 12$.Dividing the two equations: $\frac{ar^2(1 + r)}{a(1 + r)} = \frac{12}{3} \Rightarrow r^2 = 4$. Since the $G$.$P$. is increasing,$r = 2$.Substituting $r = 2$ into $a(1 + 2) = 3$,we get $3a = 3$,so $a = 1$.The terms are $1, 2, 4, 8$.Thus,$p = ab = 1 	imes 2 = 2$ and $q = cd = 4 	imes 8 = 32$.The ratio $(q + p) : (q - p) = (32 + 2) : (32 - 2) = 34 : 30 = 17 : 15$.

Let $a$ and $b$ be roots of $x^2 - 3x + p = 0$ and let $c$ and $d$ be the roots of $x^2 - 12x + q = 0$,where $a, b, c, d$ form an increasing $G$.$P$. Then the ratio of $(q + p) : (q - p)$ is equal to

Similar Questions

If the sum of an infinite geometric series is $\frac{4}{5}$ and its $1^{st}$ term is $\frac{3}{4}$, then its common ratio is

The geometric mean of $n$ positive terms $x_1, x_2, \dots, x_n$ is equal to:

If the $5^{th}$ term of a $G.P.$ is $\frac{1}{3}$ and the $9^{th}$ term is $\frac{16}{243}$,then the $4^{th}$ term will be:

Let $S_1$ be the sum of areas of the squares whose sides are parallel to the coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then,$\frac{S_1}{S_2}$ is equal to

The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module