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 $I$: $12 x^{2} + 11 x - 56 = 0$.Splitting the middle term: $12 x^{2} + 32 x - 21 x - 56 = 0$.$4 x(3 x + 8) - 7(3 x + 8) = 0$.$(4 x -

Vedclass · Accepted Answer

if $x \le y$

Vedclass · Answer

(D) For equation $I$: $12 x^{2} + 11 x - 56 = 0$.Splitting the middle term: $12 x^{2} + 32 x - 21 x - 56 = 0$.$4 x(3 x + 8) - 7(3 x + 8) = 0$.$(4 x - 7)(3 x + 8) = 0$.So,$x = \frac{7}{4} = 1.75$ or $x = -\frac{8}{3} \approx -2.67$.For equation $II$: $4 y^{2} - 15 y + 14 = 0$.Splitting the middle term: $4 y^{2} - 8 y - 7 y + 14 = 0$.$4 y(y - 2) - 7(y - 2) = 0$.$(4 y - 7)(y - 2) = 0$.So,$y = \frac{7}{4} = 1.75$ or $y = 2$.Comparing the values:$x = \{1.75, -2.67\}$ and $y = \{1.75, 2\}$.Since $1.75 \le 1.75$,$1.75 < 2$,$-2.67 < 1.75$,and $-2.67 < 2$,we conclude that $x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $12 x^{2} + 11 x - 56 = 0$
$II.$ $4 y^{2} - 15 y + 14 = 0$

Similar Questions

Divide $16$ into $2$ parts such that twice the square of the larger part exceeds the square of the smaller part by $164$.

The interval of $k$ for which the smaller root of the equation $x^2+kx-4=0$ lies in the interval $(-1, 2)$ is:

Solve the given two equations and select the correct answer from the given options.
$I. \quad x - 7\sqrt{x} + 12 = 0$
$II. \quad y - 5\sqrt{y} + 6 = 0$

Solve the given two equations and select the correct option.
$I.$ $\frac{18}{x^2} + \frac{6}{x} - \frac{12}{x^2} = \frac{8}{x^2}$
$II.$ $y^3 + 9.68 + 5.64 = 16.95$

The expression $y = ax^2 + bx + c$ has always the same sign as $c$ if

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $12 x^{2} + 11 x - 56 = 0$$II.$ $4 y^{2} - 15 y + 14 = 0$