search
Log In
0 votes
53 views

Let $p,q,r,s$ be real numbers such that $pr=2(q+s).$ Consider the equations $x^{2}+px+q=0$ and $x^{2}+rx+s=0.$ Then

  1. at least one of the equations has real roots.
  2. both these equations have real roots.
  3. neither of these equations have real roots.
  4. given data is not sufficient to arrive at any conclusion.
in Numerical Ability
recategorized by
53 views

1 Answer

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

$\underline{\mathbf{Explanation:}}$

The discriminant of the equation $\mathrm{x^2 + px + q = 0}$ is $\Delta_1 = \mathrm{p^2-4q}$

and discriminant of $\mathrm{x^2 = rx = s = 0}$ is given by $\Delta_2 = \mathrm{r^2 - 4s}$

Now, $\Delta_1 + \Delta_2 = \mathrm{p^2 - 4q + r^2 -4s = p^2 + r^2 -4(q+s)  = p^2 + r^2 -2pr = (p-r)^2} \;[\text{if}\;\mathrm{pr = 2(q+s)}]$

$\therefore \Delta_1 + \Delta_2 = \mathrm{(p-r)^2} \ge 0$

$\because$ Sum of these two $\mathbf{Discriminant}$ is $\mathbf{positive}$.

$\therefore $ At least one of them has to be $\mathbf{positive}$.

$\therefore$ At least one of the equations between $\mathrm{x^2+px+q  = 0}$ and $\mathrm{x^2 + rx + s = 0}$ must have real roots.

$\therefore \mathbf A$ is the correct option.

edited by
1

Here needs some corrections. $\Delta_1+\Delta_2=(p-r)^2$ not $(p+r)^2$

$p^2+r^2-4(q+s)=p^2+r^2-2\{2(q+r)\}=p^2+r^2-2pr=(p-r)^2$

1
Yes, thanks!

Related questions

0 votes
1 answer
1
43 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 43 views
1 vote
1 answer
2
47 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 47 views
1 vote
2 answers
3
74 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 74 views
1 vote
1 answer
4
66 views
If one root of a quadratic equation $ax^{2}+bx+c=0$ be equal to the n th power of the other, then $(ac)^{\frac{n}{n+1}}+b=0$ $(ac)^{\frac{n+1}{n}}+b=0$ $(ac^{n})^{\frac{1}{n+1}}+(a^{n}c)^{\frac{1}{n+1}}+b=0$ $(ac^\frac{1}{n+1})^{n}+(a^\frac{1}{n+1}c)^{n+1}+b=0$
asked Sep 18, 2019 in Numerical Ability gatecse 66 views
...