
If $3{p^2} = 5p + 2$ and $3{q^2} = 5q + 2$, where $p \ne q$, then $pq$ is equal to
1. $\dfrac{2}{3}$
2. $ - \dfrac{2}{3}$
3. $\dfrac{3}{2}$
4. $ - \dfrac{3}{2}$
Answer
164.1k+ views
Hint:In this question, we are given two quadratic equations $3{p^2} = 5p + 2$ and $3{q^2} = 5q + 2$. We have to calculate the value of $pq$. Also, $p$ should not be equal to $q$. Using quadratic equation calculate the value of root and put the required values in $pq$ and solve further.
Formula Used:
Let, the quadratic equation be $a{x^2} + bx + c = 0$
So, the root of the equation $a{x^2} + bx + c = 0$ will be
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Complete step by step Solution:
Given that,
There are two quadratic equations $3{p^2} = 5p + 2$ and $3{q^2} = 5q + 2$
It can be written as,
$3{p^2} - 5p - 2 = 0$ and $3{q^2} - 5q - 2 = 0$
Now, the given equation is in the same format with different variables.
Let, the equation be $3{x^2} - 5x - 2 = 0$ where $x = p = q$
Comparing $3{x^2} - 5x - 2 = 0$ with the general quadratic equation $a{x^2} + bx + c = 0$
We get, $a = 3,b = - 5,c = - 2$
Using quadratic formula to calculate the roots,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 3 \right)\left( { - 2} \right)} }}{{2\left( 3 \right)}}$
$x = \dfrac{{5 \pm \sqrt {25 + 24} }}{6}$
$x = \dfrac{{5 \pm \sqrt {49} }}{6}$
$x = \dfrac{{5 \pm 7}}{6}$
$x = \dfrac{{5 + 7}}{6},\dfrac{{5 - 7}}{6}$
$x = \dfrac{{12}}{6},\dfrac{{ - 2}}{6} \approx 2,\dfrac{{ - 1}}{3}$
As we know that, $x = p = q$
It implies that,
The values of $p$ and $q$ are $2,\dfrac{{ - 1}}{3}$.
Now, we are given that $p$ and $q$ cannot be equal.
So, there will be two ways to calculate the product of $p$ and $q$ either take $p = 2,q = \dfrac{{ - 1}}{3}$ or $q = 2,p = \dfrac{{ - 1}}{3}$. But the product in both cases will be the same.
Let, $p = 2,q = \dfrac{{ - 1}}{3}$
Therefore, $pq = 2\left( {\dfrac{{ - 1}}{3}} \right) = - \dfrac{2}{3}$
Thus, the required value of the product of $p$ and $q$ is $ - \dfrac{2}{3}$.
Hence, the correct option is (2).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
Formula Used:
Let, the quadratic equation be $a{x^2} + bx + c = 0$
So, the root of the equation $a{x^2} + bx + c = 0$ will be
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Complete step by step Solution:
Given that,
There are two quadratic equations $3{p^2} = 5p + 2$ and $3{q^2} = 5q + 2$
It can be written as,
$3{p^2} - 5p - 2 = 0$ and $3{q^2} - 5q - 2 = 0$
Now, the given equation is in the same format with different variables.
Let, the equation be $3{x^2} - 5x - 2 = 0$ where $x = p = q$
Comparing $3{x^2} - 5x - 2 = 0$ with the general quadratic equation $a{x^2} + bx + c = 0$
We get, $a = 3,b = - 5,c = - 2$
Using quadratic formula to calculate the roots,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 3 \right)\left( { - 2} \right)} }}{{2\left( 3 \right)}}$
$x = \dfrac{{5 \pm \sqrt {25 + 24} }}{6}$
$x = \dfrac{{5 \pm \sqrt {49} }}{6}$
$x = \dfrac{{5 \pm 7}}{6}$
$x = \dfrac{{5 + 7}}{6},\dfrac{{5 - 7}}{6}$
$x = \dfrac{{12}}{6},\dfrac{{ - 2}}{6} \approx 2,\dfrac{{ - 1}}{3}$
As we know that, $x = p = q$
It implies that,
The values of $p$ and $q$ are $2,\dfrac{{ - 1}}{3}$.
Now, we are given that $p$ and $q$ cannot be equal.
So, there will be two ways to calculate the product of $p$ and $q$ either take $p = 2,q = \dfrac{{ - 1}}{3}$ or $q = 2,p = \dfrac{{ - 1}}{3}$. But the product in both cases will be the same.
Let, $p = 2,q = \dfrac{{ - 1}}{3}$
Therefore, $pq = 2\left( {\dfrac{{ - 1}}{3}} \right) = - \dfrac{2}{3}$
Thus, the required value of the product of $p$ and $q$ is $ - \dfrac{2}{3}$.
Hence, the correct option is (2).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
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