(C) For equation $I$: $2x^{2} + 11x + 14 = 0$$2x^{2} + 4x + 7x + 14 = 0$$2x(x + 2) + 7(x + 2) = 0$$(2x + 7)(x + 2) = 0$$x = -3.5$ or $x = -2$For equat

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

(C) For equation $I$: $2x^{2} + 11x + 14 = 0$$2x^{2} + 4x + 7x + 14 = 0$$2x(x + 2) + 7(x + 2) = 0$$(2x + 7)(x + 2) = 0$$x = -3.5$ or $x = -2$For equation $II$: $4y^{2} + 12y + 9 = 0$$(2y)^{2} + 2(2y)(3) + 3^{2} = 0$$(2y + 3)^{2} = 0$$2y = -3$$y = -1.5$Comparing the values:$x_1 = -3.5, x_2 = -2$$y = -1.5$Since $-3.5 < -1.5$ and $-2 < -1.5$,we conclude that $x < y$.

Competitive Quantitative Aptitude Algebra QUADRATIC EQUATION

Easy

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^{2} + 11x + 14 = 0$
$II.$ $4y^{2} + 12y + 9 = 0$

A
if $x > y$
B
if $x \geq y$
C
if $x < y$
D
if $x \leq y$

Explore More

Browse All

Question Bank

All Subjects

Competitive Questions

Competitive

Quantitative Aptitude Questions

Chapter

Algebra

Subtopic

QUADRATIC EQUATION

$I. 3x^2 - 20x - 32 = 0$
$II. 2y^2 - 3y - 20 = 0$

Difficult

View Solution

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

Difficult

View Solution

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$

Medium

View Solution

If the product of the roots of the equation $(a + 1)x^2 + (2a + 3)x + (3a + 4) = 0$ is $2$,then the sum of the roots is:

Easy

View Solution

For what value of $k$ will the equation $x^2 - (3k - 1)x + 2k^2 + 2k - 11 = 0$ have equal roots?

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Solve the given two equations and select the correct answer from the given options.$I.$ $2x^{2} + 11x + 14 = 0$$II.$ $4y^{2} + 12y + 9 = 0$

Similar Questions

$I. 3x^2 - 20x - 32 = 0$$II. 2y^2 - 3y - 20 = 0$

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-25=0$$II.$ $4y^{2}=24y-35$

If the product of the roots of the equation $(a + 1)x^2 + (2a + 3)x + (3a + 4) = 0$ is $2$,then the sum of the roots is:

For what value of $k$ will the equation $x^2 - (3k - 1)x + 2k^2 + 2k - 11 = 0$ have equal roots?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^{2} + 11x + 14 = 0$
$II.$ $4y^{2} + 12y + 9 = 0$

$I. 3x^2 - 20x - 32 = 0$
$II. 2y^2 - 3y - 20 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$