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) Given equations are $9x^2 - 29x + 22 = 0$ and $y^2 - 7y + 12 = 0$.For the first equation: $9x^2 - 29x + 22 = 0$$9x^2 - 18x - 11x + 22 = 0$$9x(x -

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(C) Given equations are $9x^2 - 29x + 22 = 0$ and $y^2 - 7y + 12 = 0$.For the first equation: $9x^2 - 29x + 22 = 0$$9x^2 - 18x - 11x + 22 = 0$$9x(x - 2) - 11(x - 2) = 0$$(9x - 11)(x - 2) = 0$So,$x = 2$ or $x = 11/9 \approx 1.22$.For the second equation: $y^2 - 7y + 12 = 0$$y^2 - 4y - 3y + 12 = 0$$y(y - 4) - 3(y - 4) = 0$$(y - 3)(y - 4) = 0$So,$y = 3$ or $y = 4$.Comparing the values: $x$ values are ${1.22, 2}$ and $y$ values are ${3, 4}$.Since all values of $x$ are less than all values of $y$,we conclude $x < y$.

Solve the given two equations and select the correct option.

Similar Questions

If $ax^2 + bx + c = 0$ has real and distinct roots,$\alpha$ and $\beta$ $(\beta > \alpha)$. Further $a > 0, b < 0$ and $c < 0$ then :-

If $\lambda$ is the ratio of the roots of the quadratic equation in $x$,$3m^2x^2 + m(m - 4)x + 2 = 0$,then the least value of $m$ for which $\lambda + \frac{1}{\lambda} = 1$ is:

Solve the given two equations and provide the correct answer from the given options.
$I.$ $4x^2 - 29x + 45 = 0$
$II.$ $3y^2 - 19y + 28 = 0$

$I. \quad p^{2}-18 p+77=0$
$II. \quad 3 q^{2}-25 q+28=0$
$Quantity \, 1$: Value of $p$
$Quantity \, 2$: Value of $q$

Solve the given two equations and select the correct answer from the given options.
$I.$ $13x - 21 = 200 - 4x$
$II.$ $y = \sqrt[3]{2197}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module