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 $I$: $18 x^{2}+18 x+4=0$Divide by $2$: $9 x^{2}+9 x+2=0$$9 x^{2}+6 x+3 x+2=0$$3 x(3 x+2)+1(3 x+2)=0$$(3 x+1)(3 x+2)=0$$x = -\frac{1}{

Vedclass · Accepted Answer

if $x \geq y$

Vedclass · Answer

(B) For equation $I$: $18 x^{2}+18 x+4=0$Divide by $2$: $9 x^{2}+9 x+2=0$$9 x^{2}+6 x+3 x+2=0$$3 x(3 x+2)+1(3 x+2)=0$$(3 x+1)(3 x+2)=0$$x = -\frac{1}{3}, -\frac{2}{3}$For equation $II$: $12 y^{2}+29 y+14=0$$12 y^{2}+21 y+8 y+14=0$$3 y(4 y+7)+2(4 y+7)=0$$(3 y+2)(4 y+7)=0$$y = -\frac{2}{3}, -\frac{7}{4}$Comparing values:$x_1 = -0.33, x_2 = -0.66$$y_1 = -0.66, y_2 = -1.75$Since $x_1 > y_1$,$x_1 > y_2$,$x_2 = y_1$,and $x_2 > y_2$,we conclude $x \geq y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $18 x^{2}+18 x+4=0$
$II.$ $12 y^{2}+29 y+14=0$

Similar Questions

${x \in R : |x - 2| = x^2} = $

The value of $a (a \ge 3)$ for which the sum of the cubes of the roots of $x^2 - (a - 2)x + (a - 3) = 0$ assumes the least value is

The roots of the equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$ are given. If one of the roots is $3 + i\sqrt{6}$, find the other roots.

If $x^{2}-3x+1=0$,then the value of $\frac{x^{6}+x^{4}+x^{2}+1}{x^{3}}$ is:

If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4 < \alpha_5 < \alpha_6$,then the equation $(x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6) = 0$ has :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $18 x^{2}+18 x+4=0$$II.$ $12 y^{2}+29 y+14=0$