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

(D) For equation $I: x^{2} + 12x + 32 = 0$Factorizing the quadratic equation: $x^{2} + 8x + 4x + 32 = 0$$x(x + 8) + 4(x + 8) = 0$$(x + 4)(x + 8) = 0$T

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For equation $I: x^{2} + 12x + 32 = 0$Factorizing the quadratic equation: $x^{2} + 8x + 4x + 32 = 0$$x(x + 8) + 4(x + 8) = 0$$(x + 4)(x + 8) = 0$Thus,$x = -4$ or $x = -8$.For equation $II: 2y^{2} + 15y + 27 = 0$Factorizing the quadratic equation: $2y^{2} + 6y + 9y + 27 = 0$$2y(y + 3) + 9(y + 3) = 0$$(2y + 9)(y + 3) = 0$Thus,$y = -4.5$ or $y = -3$.Comparing the values:When $x = -4$,$x > y$ (since $-4 > -4.5$) and $x When $x = -8$,$x Since the relationship changes depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I. x^{2} + 12x + 32 = 0$
$II. 2y^{2} + 15y + 27 = 0$

Similar Questions

The number of solutions of the equation $2^x = x^2$ is

Solve the given two equations and select the correct answer from the given options:
$I. 6x^{2} + 29x + 35 = 0$
$II. 3y^{2} + 19y + 30 = 0$

The set of values of $a$ such that $x^2 - 2ax + a^2 - 6a \leqslant 0$ for all $x \in [1, 2]$ is:

If $\alpha, \beta$ are the roots of $x^2 - 2x + 4 = 0$,then $\alpha^5 + \beta^5$ is equal to

In a cubic equation,the coefficient of $x^2$ is zero and the remaining coefficients are real. If one root is $\alpha = 3 + 4i$ and the remaining roots are $\beta$ and $\gamma$,then find the value of $\alpha \beta \gamma$.

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} + 12x + 32 = 0$$II. 2y^{2} + 15y + 27 = 0$