Question

The function ‘g’ is defined by \[g(x) = 6{x^2} + x - 2\]. In the x-y plane, the graph \[y = g(x)\] crosses the x-axis at \[x = a\]. Which of the following can be the value of ‘a’?
A. \[\dfrac{1}{2}\]
B. $1$
C. \[\dfrac{3}{2}\]
D. $2$
E. \[\dfrac{5}{2}\]

Answer 1

Hint: We use the concept that the x-y plane has two axes and that any point on the x-axis has its y coordinate as zero. As the graph of the function crosses the x-axis at a point ‘a’, then the y coordinate will be zero there. Equate the function to zero and calculate the value of ‘a’. Use factorization method to find the roots of quadratic equations.

Complete step-by-step answer:
We are given the function \[g(x) = 6{x^2} + x - 2\] has a graph in the x-y plane.
We know if the graph cuts the x-axis at any point, then at that point the coordinate of y-axis will be zero.
Since graph is represented by \[y = g(x)\]
\[ \Rightarrow y = 6{x^2} + x - 2\] ……………...… (1)
We are given that the graph \[y = g(x)\] crosses the x-axis at \[x = a\]
Let \[(x,y)\]be any point on the graph of \[y = g(x)\]
So, at point \[(x,y) = (a,0)\]
Put the value of \[x = a,y = 0\] in equation (1)
\[ \Rightarrow 6{a^2} + a - 2 = 0\]
We use factorization method to find the values of ‘a’
\[ \Rightarrow 6{a^2} - 3a + 4a - 2 = 0\]
Take 3a common from first two terms and 2 common from last two terms
\[ \Rightarrow 3a(2a - 1) + 2(2a - 1) = 0\]
\[ \Rightarrow (3a + 2)(2a - 1) = 0\]
Equate each factor to zero
\[ \Rightarrow 3a + 2 = 0\]
Shift constant values to one side
\[ \Rightarrow 3a = - 2\]
Divide both sides by 3
\[ \Rightarrow a = \dfrac{{ - 2}}{3}\]............… (2)
\[ \Rightarrow 2a - 1 = 0\]
Shift constant values to one side
\[ \Rightarrow 2a = 1\]
Divide both sides by 2
\[ \Rightarrow a = \dfrac{1}{2}\]...............… (3)
From equations (2) and (3), values of ‘a’ are \[\dfrac{{ - 2}}{3}\]and\[\dfrac{1}{2}\].
Since only one option matches the answer.
\[\therefore a = \dfrac{1}{2}\]

\[\therefore \]Correct answer is option A.

Note: Alternate method:
Another method to find the roots of quadratic equation \[6{a^2} + a - 2 = 0\]
General form of quadratic equation is \[a{x^2} + bx + c = 0\] and the roots are given by the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
On comparing the equation \[6{a^2} + a - 2 = 0\] with general equation
\[a = 6,b = 1,c = - 2\]
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Put the values
\[ \Rightarrow a = \dfrac{{ - 1 \pm \sqrt {{{(1)}^2} - 4 \times 6 \times ( - 2)} }}{{2 \times 6}}\]
Write multiplication of two negative signs as positive signs
\[ \Rightarrow a = \dfrac{{ - 1 \pm \sqrt {1 + 48} }}{{12}}\]
\[ \Rightarrow a = \dfrac{{ - 1 \pm \sqrt {49} }}{{12}}\]
\[ \Rightarrow a = \dfrac{{ - 1 \pm \sqrt {{7^2}} }}{{12}}\]
Cancel square root by square power
\[ \Rightarrow a = \dfrac{{ - 1 \pm 7}}{{12}}\]
Either \[a = \dfrac{{ - 1 + 7}}{{12}}\]or\[a = \dfrac{{ - 1 - 7}}{{12}}\]
Either \[a = \dfrac{6}{{12}}\] or \[a = \dfrac{{ - 8}}{{12}}\]
Cancel same factors from numerator and denominator
Either \[a = \dfrac{1}{2}\] or \[a = \dfrac{{ - 2}}{3}\]
Since only one option matches the answer.
\[\therefore a = \dfrac{1}{2}\]
\[\therefore \]Correct option is A.