Question

For a given simultaneous equation mark the correct answer.
1) $6{x^2} - 49x + 99 = 0$
2) $5{y^2} + 17y + 14 = 0$
A) If $x < y$
B) If $x > y$
C) If $x \leqslant y$
D) If $x \geqslant y$
E) If $x = y$, or relationship between $x$ and $y$ can’t be determined.

Answer 1

Hint:
Here, we are required to find the relationship between the two variables $x$ and $y$ when we are given two quadratic equations. Hence, we will solve these equations individually by doing a middle term split and hence, find the respective values of $x$ and $y$. Comparing these values, we will be able to find the required relationship between the two variables.

Complete step by step solution:
Given two equations are:
$6{x^2} - 49x + 99 = 0$
And $5{y^2} + 17y + 14 = 0$
Now, we will solve the first quadratic equation using middle term splitting,
Hence, $6{x^2} - 49x + 99 = 0$ can be written as:
$ \Rightarrow 6{x^2} - 27x - 22x + 99 = 0$
$ \Rightarrow 3x\left( {2x - 9} \right) - 11\left( {2x - 9} \right) = 0$
Taking brackets common, we get,
$ \Rightarrow \left( {3x - 11} \right)\left( {2x - 9} \right) = 0$
Therefore, either $3x - 11 = 0$
$ \Rightarrow x = \dfrac{{11}}{3}$
Or $2x - 9 = 0$
$ \Rightarrow x = \dfrac{9}{2}$
Hence, the values of $x$ are $x = \dfrac{{11}}{3},\dfrac{9}{2}$
Converting these fractions into decimals, we get,
$x = 3.67,4.5$
Now, we will solve the second quadratic equation using middle term splitting,
Hence, $5{y^2} + 17y + 14 = 0$ can be written as:
$ \Rightarrow 5{y^2} + 10y + 7y + 14 = 0$
$ \Rightarrow 5y\left( {y + 2} \right) + 7\left( {y + 2} \right) = 0$
Taking brackets common, we get,
$ \Rightarrow \left( {5y + 7} \right)\left( {y + 2} \right) = 0$
Therefore, either $5y + 7 = 0$
$ \Rightarrow y = \dfrac{{ - 7}}{5}$
Or $y + 2 = 0$
$ \Rightarrow y = - 2$
Hence, the values of $y$ are $y = \dfrac{{ - 7}}{5}, - 2$
Converting these fractions into decimals, we get,
$y = - 1.4, - 2$
Now, comparing the values of $x$ and $y$, we can clearly see that both the values of $x$ are positive as $x = 3.67,4.5$ whereas, both the values of $y$ are negative as $y = - 1.4, - 2$
Hence, there can’t be any case in which $y$ will be greater than or even equal to $x$

Therefore, the value of $x$ will always remain greater than the value of $y$
Thus, $x > y$
Hence, option B is the correct answer.

Note:
An alternate way of solving this question is by using the quadratic formula, according to which, we compare the given equations by the general formula of a quadratic equation i.e. $a{x^2} + bx + c = 0$ and then, we can find the roots of this equation by using the quadratic formula , $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Hence, for the first equation $6{x^2} - 49x + 99 = 0$, comparing with $a{x^2} + bx + c = 0$
We get,
$a = 6,b = - 49,c = 99$
Substituting these values we get,
$x = \dfrac{{49 \pm \sqrt {{{\left( { - 49} \right)}^2} - 4\left( 6 \right)\left( {99} \right)} }}{{2\left( 6 \right)}} = \dfrac{{49 \pm \sqrt {2401 - 2376} }}{{12}} = \dfrac{{49 \pm \sqrt {22} }}{{12}}$
$ \Rightarrow x = \dfrac{{49 + 4.7}}{{12}},\dfrac{{49 - 4.7}}{{12}}$
Solving further,
$ \Rightarrow x = 4.5,3.67$
Hence, we get the same value of $x$
Similarly, just by substituting the coefficients, we will get the same value of $y$
And hence, we will find that $x > y$
Thus, we can use either of the two methods to solve the quadratic equations and then compare them accordingly.