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) Step $1$: Solve for $x$.$x = \sqrt{172}$. Since $13^2 = 169$ and $14^2 = 196$,the value of $x$ is approximately $13.11$.Step $2$: Solve the quadra

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(B) Step $1$: Solve for $x$.$x = \sqrt{172}$. Since $13^2 = 169$ and $14^2 = 196$,the value of $x$ is approximately $13.11$.Step $2$: Solve the quadratic equation for $y$.$y^{2} - 29y + 210 = 0$.We need two numbers that multiply to $210$ and add up to $29$. These numbers are $14$ and $15$ ($14 	imes 15 = 210$ and $14 + 15 = 29$).So,$(y - 14)(y - 15) = 0$.This gives $y = 14$ or $y = 15$.Step $3$: Compare $x$ and $y$.Since $x \approx 13.11$ and $y$ is either $14$ or $15$,it is clear that $x Therefore,the correct option is $B$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt{172}$
$II.$ $y^{2} - 29y + 210 = 0$

Similar Questions

If the roots of the equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ are equal,then $a, b, c$ are in

If the sum of the roots of the equation $ax^2 + bx + c = 0$ is equal to the sum of their squares,then:

If $\alpha$ and $\beta$ are two real numbers satisfying $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$,then the value of $\alpha \beta$ is:

Solve the given two equations and select the correct answer from the given options.
$I. \quad 6x^{2} - 49x + 99 = 0$
$II. \quad 5y^{2} + 17y + 14 = 0$

The value of $p$ so that both the roots of the equation $(p - 5)x^2 - 2px + (p - 4) = 0$ are positive,one is less than $2$ and other is lying between $2$ and $3$,lies in the interval:

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 = \sqrt{172}$$II.$ $y^{2} - 29y + 210 = 0$