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 $I$: $x^{2}-25=0 \Rightarrow x^{2}=25 \Rightarrow x = 5, -5$.For equation $II$: $4y^{2}-24y+35=0$.Using the quadratic formula or fact

Vedclass · Accepted Answer

if $x = y$ or relationship between $x$ and $y$ cannot be established.

Vedclass · Answer

(D) For equation $I$: $x^{2}-25=0 \Rightarrow x^{2}=25 \Rightarrow x = 5, -5$.For equation $II$: $4y^{2}-24y+35=0$.Using the quadratic formula or factorization: $4y^{2}-14y-10y+35=0$.$2y(2y-7)-5(2y-7)=0 \Rightarrow (2y-5)(2y-7)=0$.Thus,$y = 2.5$ or $y = 3.5$.Comparing the values:If $x = 5$,then $x > 2.5$ and $x > 3.5$ $(x > y)$.If $x = -5$,then $x Since we get different relationships ($x > y$ and $x < y$),the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$

Similar Questions

The value of $a (a \ge 3)$ for which the sum of the cubes of the roots of $x^2 - (a - 2)x + (a - 3) = 0$ assumes the least value is

Solve the given two equations and select the correct option.
$I.$ $2x^2 + 23x + 63 = 0$
$II.$ $4y^2 + 19y + 21 = 0$

If $(3x - 2y) : (2x + 3y) = 5 : 6$,then one of the values of $\left(\frac{\sqrt[3]{x} + \sqrt[3]{y}}{\sqrt[3]{x} - \sqrt[3]{y}}\right)^{2}$ is

One root of the quadratic equation $x^{2}-5x+6=0$ is $3$. Find the other root.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-25=0$$II.$ $4y^{2}=24y-35$