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

(C) Given equations are $9x^2 - 29x + 22 = 0$ and $y^2 - 7y + 12 = 0$.For the first equation: $9x^2 - 29x + 22 = 0$$9x^2 - 18x - 11x + 22 = 0$$9x(x -

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(C) Given equations are $9x^2 - 29x + 22 = 0$ and $y^2 - 7y + 12 = 0$.For the first equation: $9x^2 - 29x + 22 = 0$$9x^2 - 18x - 11x + 22 = 0$$9x(x - 2) - 11(x - 2) = 0$$(9x - 11)(x - 2) = 0$So,$x = 2$ or $x = 11/9 \approx 1.22$.For the second equation: $y^2 - 7y + 12 = 0$$y^2 - 4y - 3y + 12 = 0$$y(y - 4) - 3(y - 4) = 0$$(y - 3)(y - 4) = 0$So,$y = 3$ or $y = 4$.Comparing the values: $x$ values are ${1.22, 2}$ and $y$ values are ${3, 4}$.Since all values of $x$ are less than all values of $y$,we conclude $x < y$.

Solve the given two equations and select the correct option.

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I. \quad x - 7\sqrt{x} + 12 = 0$
$II. \quad y - 5\sqrt{y} + 6 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} = 14641$
$II.$ $y = \sqrt{14641}$

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

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 $3 + 4i$ is a root of the equation ${x^2} + px + q = 0$ ($p, q$ are real numbers),then

Vedclass Test Series

Exam Paper Generator

Online Exam Module