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 $I$: $12x^2 - 47x + 40 = 0$Splitting the middle term: $12x^2 - 32x - 15x + 40 = 0$$4x(3x - 8) - 5(3x - 8) = 0$$(4x - 5)(3x - 8) = 0$S

Vedclass · Accepted Answer

if $x \ge y$

Vedclass · Answer

(B) For equation $I$: $12x^2 - 47x + 40 = 0$Splitting the middle term: $12x^2 - 32x - 15x + 40 = 0$$4x(3x - 8) - 5(3x - 8) = 0$$(4x - 5)(3x - 8) = 0$So,$x = \frac{5}{4} = 1.25$ and $x = \frac{8}{3} \approx 2.67$.For equation $II$: $4y^2 + 3y - 10 = 0$Splitting the middle term: $4y^2 + 8y - 5y - 10 = 0$$4y(y + 2) - 5(y + 2) = 0$$(4y - 5)(y + 2) = 0$So,$y = \frac{5}{4} = 1.25$ and $y = -2$.Comparing the values:$x_1 = 1.25, x_2 = 2.67$$y_1 = 1.25, y_2 = -2$Since $x_1 = y_1$,$x_2 > y_1$,and $x_1, x_2 > y_2$,we conclude that $x \ge y$.

Solve the given two equations and select the correct option.
$I.$ $12x^2 - 47x + 40 = 0$
$II.$ $4y^2 + 3y - 10 = 0$

Similar Questions

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is always non-negative for $x > 0$,then the minimum value of $m$ must be

The equation whose roots are reciprocal of the roots of the equation $3x^2 - 20x + 17 = 0$ is

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

The number of solutions for the equation ${x^2} - 5|x| + 6 = 0$ is

The roots of the given equation $(p - q){x^2} + (q - r)x + (r - p) = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct option.$I.$ $12x^2 - 47x + 40 = 0$$II.$ $4y^2 + 3y - 10 = 0$