search
Log In
1 vote
55 views

If one root of a quadratic equation $ax^{2}+bx+c=0$ be equal to the n th power of the other, then

  1. $(ac)^{\frac{n}{n+1}}+b=0$
  2. $(ac)^{\frac{n+1}{n}}+b=0$
  3. $(ac^{n})^{\frac{1}{n+1}}+(a^{n}c)^{\frac{1}{n+1}}+b=0$
  4. $(ac^\frac{1}{n+1})^{n}+(a^\frac{1}{n+1}c)^{n+1}+b=0$
in Numerical Ability
recategorized by
55 views

1 Answer

1 vote
$\underline{\mathbf{Answer:3}}$

$\underline{\mathbf{Solution:}}$

Let $\alpha$ and $\beta$ are the roots of the equation $\mathrm{ax^2 + bx + c = 0}$

then

$\mathrm {B = \alpha^n}\;\text{[Given]}$

Also we know that:

$\text{Sum of roots} = -\dfrac{b}{a}$

$\text{Product of roots} = \mathrm{\dfrac{c}{a}}$

$\therefore \alpha = \mathrm{\frac{c}{a}}\Rightarrow \alpha\times \alpha^n =\alpha^{n+1}= \dfrac{c}{a} \Rightarrow \alpha = (\frac{c}{a})^{\frac{1}{n+1}}$

then,

$\alpha + \beta = -\frac{b}{a} \Rightarrow \alpha + \alpha^n= -\frac{b}{a}$

Now we can write the quadratic equation $\mathrm{ax^2+bx+c = 0}$ using one of the root $\alpha$ as:

$\mathrm{a\alpha^2 + b\alpha + c = 0}$

$\Rightarrow \mathrm{a\alpha +b + \frac{c}{\alpha}}$

$\mathrm{\Rightarrow \frac{c}{\alpha} + a\alpha + b = 0}$

$\mathrm{\dfrac{c}{(\frac{c}{a})^{\frac{1}{n+1}}}+a\left (\frac{c}{a} \right )^{\frac{1}{n+1}}+b = 0}$

$\mathrm{c \left (\frac{a}{c} \right )^{\frac{1}{n+1}}+a\left ( \frac{c}{a}\right)^{\frac{1}{n+1}}+b = 0}$

$\mathrm{\left ( \frac{c^{n+1}a}{c}\right)^{\frac{1}{n+1}}+ \left ( \frac{a^{n+1}c}{a}\right)^{\frac{1}{n+1}} + b = 0}$

$\mathrm{\Rightarrow (ac^n)^{\frac{1}{n+1}} + (a^nc)^{\frac{1}{n+1}} + b = 0}$

$\therefore \;\mathrm{Option}\; 3$ is the correct answer.

edited by
0
Nice answer. But it has some typing mistakes. Please correct those.

It will be $\alpha \times \alpha^n=\frac{c}{a}\Rightarrow \alpha^{n+1}=\frac{c}{a}\Rightarrow \alpha=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}$. Then do some calculations.
0
I think that is correct.

Can you please checkout once?
1
Yeah, you are correct.

That's a typing error.
1
PS

Corrected!
0
Yeah. It's now correct.

Related questions

0 votes
1 answer
1
39 views
Let $x^{2}-2(4k-1)x+15k^{2}-2k-7>0$ for any real value of $x$. Then the integer value of $k$ is $2$ $4$ $3$ $1$
asked Sep 18, 2019 in Numerical Ability gatecse 39 views
1 vote
1 answer
2
43 views
If $\alpha$ and $\beta$ be the roots of the equation $x^{2}+3x+4=0,$ then the equation with roots $(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$ is $x^{2}+2x+63=0$ $x^{2}-63x+2=0$ $x^{2}-2x-63=0$ None of these
asked Sep 18, 2019 in Numerical Ability gatecse 43 views
1 vote
2 answers
3
61 views
If $r$ be the ratio of the roots of the equation $ax^{2}+bx+c=0,$ then $\frac{r}{b}=\frac{r+1}{ac}$ $\frac{r+1}{b}=\frac{r}{ac}$ $\frac{(r+1)^{2}}{r}=\frac{b^{2}}{ac}$ $\left(\frac{r}{b}\right)^{2}=\frac{r+1}{ac}$
asked Sep 18, 2019 in Numerical Ability gatecse 61 views
1 vote
0 answers
4
49 views
The condition that ensures that the roots of the equation $x^{3}-px^{2}+qx-r=0$ are in H.P. is $r^{2}-9pqr+q^{3}=0$ $27r^{2}-9pqr+3q^{3}=0$ $3r^{3}-27pqr-9q^{3}=0$ $27r^{2}-9pqr+2q^{3}=0$
asked Sep 18, 2019 in Numerical Ability gatecse 49 views
...