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

(B) For equation $I$: $x^{2}-9x+20=0$.We need two numbers whose product is $20$ and sum is $9$. These are $5$ and $4$.So,$(x-5)(x-4)=0$,which gives $x

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(B) For equation $I$: $x^{2}-9x+20=0$.We need two numbers whose product is $20$ and sum is $9$. These are $5$ and $4$.So,$(x-5)(x-4)=0$,which gives $x = 5$ or $x = 4$.For equation $II$: $y^{2}-13y+42=0$.We need two numbers whose product is $42$ and sum is $13$. These are $7$ and $6$.So,$(y-7)(y-6)=0$,which gives $y = 7$ or $y = 6$.Comparing the values:If $x=4$,then $y=7$ $(x If $x=5$,then $y=7$ $(x In all cases,$x < y$.

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

Similar Questions

If $2 + i\sqrt{3}$ is a root of the equation $x^2 + px + q = 0$,where $p$ and $q$ are real,then $(p, q) = $

If $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 2 = 0$,then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

The roots of the equation $x^2 - \sqrt{13}x + 1 = 0$ are:

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+3ax+2a^{2}=0$ and if $\alpha^{2}+\beta^{2}=5$,find the value of $a$.

Solve the given two equations and provide the correct answer from the given options.
$I.$ $4x^2 - 29x + 45 = 0$
$II.$ $3y^2 - 19y + 28 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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