What will be the integral value of $x$ satisfying \[6{x^2} - 7x + 2 = 0\] ?
A. -2
B. -3
C. 2
D. none

Answer Verified Verified
Hint: We will first solve the given quadratic equation by the quadratic formula and then we will check the given options that which integral value satisfies the given quadratic equation.

Complete step-by-step answer-
The quadratic equation is \[6{x^2} - 7x + 2 = 0\]. Now, we will find the roots of this equation. The roots are the values of $x$ that satisfy the given quadratic equation.
Now, we will find roots through discriminant formula which is also known as Sridharacharya formula. Discriminant formula for finding roots of a quadratic equation \[a{x^2} + bx + c = 0\] is,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where \[a\] represents the coefficient of \[{x^2}\], \[b\] represents the coefficient of $x$ and \[c\] represents the constant term. Now, solving the given quadratic equation using this formula,
Where \[a\] = 6, \[b\] = -7, \[c\] = 2. On solving,
\[ \Rightarrow \] \[x = \dfrac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(6)(2)} }}{{2(6)}}\]
\[ \Rightarrow \] \[x = \dfrac{{ - ( - 7) \pm \sqrt {49 - 48} }}{{12}}\]
\[ \Rightarrow \] \[x = \dfrac{{7 \pm 1}}{{12}}\]
Now, first taking the positive sign, we get
\[x = \dfrac{{7 + 1}}{{12}} = \dfrac{8}{{12}} = \dfrac{2}{3}\]
Taking the negative sign, we get
\[x = \dfrac{{7 - 1}}{{12}} = \dfrac{6}{{12}} = \dfrac{1}{2}\]

So, the values of $x$ satisfying the given quadratic equation are \[\dfrac{2}{3}\]and \[\dfrac{1}{2}\]. Now, looking at the options given we can clearly see that there is no value which satisfies the given equation.
So, the correct answer is none i.e. option (D).

Note: Another easy method for finding the roots is the middle – term split method. Although the middle – term split method is easy and used in the majority of the questions but, in difficult questions discriminant formula is used to find the roots easily. In such types of questions applying the formula correctly will give you a correct answer free of errors.
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