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 $I: x^{2}-19x+84=0$We need two numbers whose product is $84$ and sum is $19$. These numbers are $12$ and $7$.$x^{2}-12x-7x+84=0$$x(x-

Vedclass · Accepted Answer

if $x \le y$

Vedclass · Answer

(D) For equation $I: x^{2}-19x+84=0$We need two numbers whose product is $84$ and sum is $19$. These numbers are $12$ and $7$.$x^{2}-12x-7x+84=0$$x(x-12)-7(x-12)=0$$(x-12)(x-7)=0$So,$x = 12$ or $x = 7$.For equation $II: y^{2}-25y+156=0$We need two numbers whose product is $156$ and sum is $25$. These numbers are $13$ and $12$.$y^{2}-13y-12y+156=0$$y(y-13)-12(y-13)=0$$(y-13)(y-12)=0$So,$y = 13$ or $y = 12$.Comparing the values:If $x=12$,then $y=13$ $(x If $x=7$,then $y=13$ $(x In all cases,$x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I. x^{2}-19x+84=0$
$II. y^{2}-25y+156=0$

Similar Questions

The number of roots of the quadratic equation $8\sec^2 \theta - 6\sec \theta + 1 = 0$ is

Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + px + p^3 = 0$ $(p \neq 0)$. If $(\alpha, \beta)$ is a point on the parabola $y^2 = x$,then the roots of the quadratic equation are:

The equation $\sqrt{3 x^{2}+x+5}=x-3$,where $x$ is real,has

If one root of the equation ${x^2} + px + q = 0$ is $2 + \sqrt{3}$,then the values of $p$ and $q$ are:

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are

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}-19x+84=0$$II. y^{2}-25y+156=0$