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 $I: 3x^2 - 4x - 32 = 0$Using the quadratic formula or factorization: $3x^2 - 12x + 8x - 32 = 0$$3x(x - 4) + 8(x - 4) = 0$$(3x + 8)(x

Vedclass · Accepted Answer

if $x \le y$

Vedclass · Answer

(D) For equation $I: 3x^2 - 4x - 32 = 0$Using the quadratic formula or factorization: $3x^2 - 12x + 8x - 32 = 0$$3x(x - 4) + 8(x - 4) = 0$$(3x + 8)(x - 4) = 0$So,$x = 4$ or $x = -8/3 \approx -2.67$.For equation $II: 2y^2 - 17y + 36 = 0$Factorizing: $2y^2 - 9y - 8y + 36 = 0$$y(2y - 9) - 4(2y - 9) = 0$$(y - 4)(2y - 9) = 0$So,$y = 4$ or $y = 9/2 = 4.5$.Comparing the values:$x = \{4, -2.67\}$$y = \{4, 4.5\}$Since $4 \le 4$,$4 \le 4.5$,$-2.67 \le 4$,and $-2.67 \le 4.5$,we conclude that $x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I. 3x^2 - 4x - 32 = 0$
$II. 2y^2 - 17y + 36 = 0$

Similar Questions

If $x+\frac{1}{x}=3,$ then the value of $x^{8}+\frac{1}{x^{8}}$ will be

Solve the given two equations and select the correct answer from the given options.
$I. 2x^2 + 3x + 1 = 0$
$II. 12y^2 + 7y + 1 = 0$

For what value of $\lambda$ is the sum of the squares of the roots of ${x^2} + (2 + \lambda )x - \frac{1}{2}(1 + \lambda ) = 0$ minimum?

If $\sin A, \sin B, \cos A$ are in $G.P.$,then the roots of ${x^2} + 2x \cot B + 1 = 0$ are always

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 35x + 2 = 0$,then the value of $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I. 3x^2 - 4x - 32 = 0$$II. 2y^2 - 17y + 36 = 0$