Solve the given two equations and select the | vedclass

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

Question

(D) For the first equation $x^{2}+5x+6=0$:Factoring the quadratic,we get $(x+2)(x+3)=0$.Thus,the roots are $x_1 = -2$ and $x_2 = -3$.For the second eq

Vedclass · Accepted Answer

if $x \le y$

Vedclass · Answer

(D) For the first equation $x^{2}+5x+6=0$:Factoring the quadratic,we get $(x+2)(x+3)=0$.Thus,the roots are $x_1 = -2$ and $x_2 = -3$.For the second equation $y^{2}+3y+2=0$:Factoring the quadratic,we get $(y+1)(y+2)=0$.Thus,the roots are $y_1 = -1$ and $y_2 = -2$.Comparing the roots:For $x = -2$,we have $y = -1$ (where $x For $x = -3$,we have $y = -1$ (where $x In all cases,$x \le y$.

Solve the given two equations and select the correct option.
$I.$ $x^{2}+5x+6=0$
$II.$ $y^{2}+3y+2=0$

Similar Questions

If $\alpha, \beta$ are the roots of $x^2 - 3x + a = 0, a \in R$ and $\alpha < 1 < \beta$,then:

The value of $k$ for which the equation $(K - 2)x^2 + 8x + K + 4 = 0$ has both roots real,distinct and negative is

If a root of the given equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ is $1$,then the other root will be

If the roots of the equations $x^2 - bx + c = 0$ and $x^2 - cx + b = 0$ differ by the same quantity, then $b + c$ is equal to

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+20=9x$
$II.$ $y^{2}+42=13y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct option.$I.$ $x^{2}+5x+6=0$$II.$ $y^{2}+3y+2=0$