If the difference of the roots of the equatio | vedclass

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

Question

(A) For the equation $x^2-7x+10=0$,the roots are $x=2$ and $x=5$. The difference of the roots is $|5-2|=3$.For the equation $x^2-17x+k=0$,let the root

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) For the equation $x^2-7x+10=0$,the roots are $x=2$ and $x=5$. The difference of the roots is $|5-2|=3$.For the equation $x^2-17x+k=0$,let the roots be $\alpha$ and $\beta$. The difference of the roots is $|\alpha-\beta|=3$.We know that $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.Substituting the values,we get $3^2 = (17)^2 - 4k$.$9 = 289 - 4k$.$4k = 289 - 9 = 280$.$k = 70$.The divisors of $70$ are $1, 2, 5, 7, 10, 14, 35, 70$.Comparing with the given options,$14$ is a divisor of $70$.

If the difference of the roots of the equation $x^2-7x+10=0$ is same as the difference of the roots of the equation $x^2-17x+k=0$,then a divisor of $k$ is

Similar Questions

The equation whose roots are squares of the roots of $x^4-2 x^3+6 x-21=0$ is

The roots of the equation $x^{2/3} + x^{1/3} - 2 = 0$ are

Let $m$ and $n$ be two integers such that $0 \leq m \leq 10$ and $0 \leq n \leq 10$. Then,the number of ordered pairs $(m, n)$ such that $x^2+m x+n=0$ has real roots is

One root of the given equation $2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$ is

Which of the following conditions implies that the roots of the equation $\frac{1}{4}x^2 + bx + c = 0$ are integers?

Vedclass Test Series

Exam Paper Generator

Online Exam Module