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

(C) For equation $I$: $15 x^{2}-41 x+14=0$We factorize the quadratic equation: $15 x^{2}-35 x-6 x+14=0$$5 x(3 x-7)-2(3 x-7)=0$$(5 x-2)(3 x-7)=0$So,$x

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(C) For equation $I$: $15 x^{2}-41 x+14=0$We factorize the quadratic equation: $15 x^{2}-35 x-6 x+14=0$$5 x(3 x-7)-2(3 x-7)=0$$(5 x-2)(3 x-7)=0$So,$x = \frac{2}{5} = 0.4$ or $x = \frac{7}{3} \approx 2.33$.For equation $II$: $2 y^{2}-13 y+20=0$We factorize the quadratic equation: $2 y^{2}-8 y-5 y+20=0$$2 y(y-4)-5(y-4)=0$$(2 y-5)(y-4)=0$So,$y = \frac{5}{2} = 2.5$ or $y = 4$.Comparing the values:$x_1 = 0.4, x_2 = 2.33$$y_1 = 2.5, y_2 = 4$Since both values of $x$ are less than both values of $y$ $(0.4 < 2.5, 0.4 < 4, 2.33 < 2.5, 2.33 < 4)$,we conclude that $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $15 x^{2}-41 x+14=0$
$II.$ $2 y^{2}-13 y+20=0$

Similar Questions

Let $a$,$b$,and $c$ be the roots of the equation $x^3 + 8x + 1 = 0$. Then the value of $\frac{bc}{(8b + 1)(8c + 1)} + \frac{ac}{(8a + 1)(8c + 1)} + \frac{ab}{(8a + 1)(8b + 1)}$ is equal to:

Solve the given two equations and select the correct option.
$I.$ $7x + 3y = 26$
$II.$ $2x + 17y = -41$

If both the roots of the quadratic equation ${x^2} - 2kx + {k^2} + k - 5 = 0$ are less than $5$,then $k$ lies in the interval

If $\left\{\frac{1}{2}(a-b)\right\}^{2}+a b=p(a+b)^{2},$ then the value of $p$ is (assume that $a \neq-b).$

The number of real roots of the equation $5 + |2^x - 1| = 2^x(2^x - 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $15 x^{2}-41 x+14=0$$II.$ $2 y^{2}-13 y+20=0$