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 $|x - 2| + |x - 3| = 7$,then $x =$

If $x = a^{1/2} + a^{-1/2}$ and $y = a^{1/2} - a^{-1/2}$,then find the value of $(x^4 - x^2 y^2 - 1) + (y^4 - x^2 y^2 + 1)$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} = 64$
$II.$ $2y^{2} + 25y + 72 = 0$

If the inequality $kx^2 - 2x + k \geq 0$ holds good for at least one real $x$,then the complete set of values of $k$ is

Find the value$(s)$ of $P$ for which the given quadratic equation $3x^{2}-5x+P=0$ has 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. 3x^2 - 4x - 32 = 0$$II. 2y^2 - 17y + 36 = 0$