Question

If the roots of the equation \[6{x^2} - 7x + k = 0\] are rational, then \[k\] is equal to
(a) 5
(b) 5, 6
(c) 9
(d) 1, 2

Answer 1

Hint: Here, we need to find the value of \[k\]. We will use the formula for discriminant to find the discriminant. Then we will substitute the values from the option in the equation of discriminant. We will then use the property of discriminant and compare it with obtained values of discriminant. If the value from the option satisfies the discriminant, then that value is the correct value of \[k\].

Formula Used:
We will use the formula of the discriminant of a quadratic equation of the form \[a{x^2} + bx + c = 0\] is given by \[D = {b^2} - 4ac\].

Complete step-by-step answer:
First, we will find the discriminant of a quadratic equation \[6{x^2} - 7x + k = 0\].
The discriminant of a quadratic equation of the form \[a{x^2} + bx + c = 0\] is given by \[D = {b^2} - 4ac\].
Comparing the equations \[6{x^2} - 7x + k = 0\] and \[a{x^2} + bx + c = 0\], we get
\[a = 6\], \[b = - 7\], and \[c = k\]
Substituting \[a = 6\], \[b = - 7\], and \[c = k\] in the formula \[D = {b^2} - 4ac\], we get
\[ \Rightarrow D = {\left( { - 7} \right)^2} - 4\left( 6 \right)\left( k \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 49 - 24k\]
We know that:
If \[D > 0\], then the roots of the quadratic equation are real and unequal.
If \[D = 0\], then the roots of the quadratic equation are real and equal.
If \[D < 0\], then the roots of the quadratic equation are not real, that is complex.
The roots of the equation are rational. This means that the roots are real.
Thus, we can say \[D \ge 0\].
Also, it is given that the roots are rational. Thus, \[D\] is a perfect square.
Now, we will check the given options and find the correct value of \[k\] such that \[D \ge 0\] and \[D\] is a perfect square.
Substituting \[k = 5\] in the equation \[D = 49 - 24k\], we get
\[ \Rightarrow D = 49 - 24\left( 5 \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 49 - 120 = - 71\]
This is not possible since \[D \ge 0\] because the roots are rational.
Thus, option (a) is incorrect.
Substituting \[k = 6\] in the equation \[D = 49 - 24k\], we get
\[ \Rightarrow D = 49 - 24\left( 6 \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 49 - 144 = - 95\]
This is not possible since \[D \ge 0\] because the roots are rational.
Thus, option (b) is incorrect.
Substituting \[k = 9\] in the equation \[D = 49 - 24k\], we get
\[ \Rightarrow D = 49 - 24\left( 9 \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 49 - 216 = - 167\]
This is not possible since \[D \ge 0\] because the roots are rational.
Thus, option (c) is incorrect.
Substituting \[k = 1\] in the equation \[D = 49 - 24k\], we get
\[ \Rightarrow D = 49 - 24\left( 1 \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 49 - 24 = 25\]
Here, \[D \ge 0\].
Substituting \[k = 2\] in the equation \[D = 49 - 24k\], we get
\[ \Rightarrow D = 49 - 24\left( 2 \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 49 - 48 = 1\]
Here, \[D \ge 0\].
The numbers 25 and 1 are the square of the natural number 5 and 1 respectively. Hence, 25 and 1 are perfect squares.
Therefore, the possible values of \[k\] are 1 and 2.
Thus, the correct option is option (d).

Note: We used the term “quadratic equation” in our solution. A quadratic equation is an equation of degree 2. It is of the form \[a{x^2} + bx + c = 0\], where \[a\] is not equal to 0. A quadratic equation has 2 solutions.
We used the term ‘perfect square’ in the solution. The square of any integer is called a perfect square. Perfect squares are the squares of integers, and not the squares of any real number. For example: the square of \[5.5\] is \[30.25\]. However, since \[5.5\] is not an integer, \[30.25\] is not a perfect square.