Solve the given two equations and provide the | vedclass

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

Question

(D) For equation $I$: $x^{2}+14x+49=0$This can be written as $(x+7)^{2}=0$.Therefore,$x = -7$.For equation $II$: $y^{2}+9y=0$Factoring out $y$,we get

Vedclass · Accepted Answer

if $x = y$ or relationship between $x$ and $y$ cannot be established.

Vedclass · Answer

(D) For equation $I$: $x^{2}+14x+49=0$This can be written as $(x+7)^{2}=0$.Therefore,$x = -7$.For equation $II$: $y^{2}+9y=0$Factoring out $y$,we get $y(y+9)=0$.Therefore,$y = 0$ or $y = -9$.Comparing the values:If $x = -7$ and $y = 0$,then $x If $x = -7$ and $y = -9$,then $x > y$.Since the relationship changes depending on the value of $y$,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and provide the correct answer from the given options.
$I.$ $x^{2}+14x+49=0$
$II.$ $y^{2}+9y=0$

Similar Questions

If the graph of $y = ax^2 + bx + c$ $(a, b, c \in R)$ is as shown in the figure,where $D = b^2 - 4ac$,which of the following is incorrect?

If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

If $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2},$ then the relation between $a, b$ and $c$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-16x+63=0$
$II.$ $y^{2}-2y-35=0$

If $\left(\frac{p^{-1} q^{2}}{p^{3} q^{-2}}\right)^{\frac{1}{3}}+\left(\frac{p^{5} q^{-3}}{p^{-2} q^{3}}\right)^{\frac{1}{3}}=p^{a} q^{b},$ then the value of $a+b,$ where $p$ and $q$ are different positive positive primes,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and provide the correct answer from the given options.$I.$ $x^{2}+14x+49=0$$II.$ $y^{2}+9y=0$