Question

Draw the graph of $y=2{{x}^{2}}$ and hence solve $2{{x}^{2}}+x-6=0$ .

Answer 1

Hint: Any year which has 366 days is called a leap year. Non-leap year contains 365 days. We have to find 80% of 365 days. We will write this mathematically as $80%\times 365$ . Then, we have to convert 80% into its number form by dividing 80 by 100 and simplify.

Complete step by step answer:
We have to draw the graph of $y=2{{x}^{2}}$ and hence solve $2{{x}^{2}}+x-6=0$ . Let us first draw the graph of $y=2{{x}^{2}}$ by substituting different values of x and find the corresponding values of y.
Let us consider when $x=-3,-2,1,0,1,2,3$ .
When $x=-3$ , we can find y as
$y=2\times {{\left( -3 \right)}^{2}}=2\times 9=18$
When $x=-2$ , we can find y as
$y=2\times {{\left( -2 \right)}^{2}}=2\times 4=8$
When $x=-1$ ,
$y=2\times {{\left( -1 \right)}^{2}}=2\times 1=2$
When $x=0$ ,
$y=2\times {{\left( 0 \right)}^{2}}=0$
Similarly, we have to find the for $x=1,2,3$ .
We can tabulate the corresponding y values as follows.

x	-3	-2	-1	0	1	2	3
$y=2{{x}^{2}}$	18	8	2	0	2	4	18

Now, let us plot the points $\left( -3,18 \right),\left( -2,8 \right),\left( -1,2 \right),\left( 0,0 \right),\left( 1,2 \right),\left( 2,8 \right),\left( 3,18 \right)$ .

Now, let us consider the equation $2{{x}^{2}}+x-6=0...\left( i \right)$ .
We are given that $y=2{{x}^{2}}$ . Let us substitute this value in the equation (i).
$\Rightarrow y+x-6=0$
We have to take all the terms other than y to the RHS.
$\Rightarrow y=6-x$
We have to substitute different values of x and find the corresponding values of y.
Let us consider $x=-3$ .
$\Rightarrow y=6-\left( -3 \right)=6+3=9$
When $x=-2$ ,
$\Rightarrow y=6-\left( -2 \right)=6+2=8$
When $x=-1$ ,
$\Rightarrow y=6-\left( -1 \right)=6+1=7$
When $x=0$ ,
$\Rightarrow y=6-\left( 0 \right)=6-0=6$
Similarly, we have to find the for $x=1,2,3$ .
We can tabulate the corresponding y values as follows.

x	-3	-2	-1	0	1	2	3
$y=6-x$	9	8	7	6	5	4	3

Now, let us plot the points $\left( -3,9 \right),\left( -2,8 \right),\left( -1,7 \right),\left( 0,6 \right),\left( 1,5 \right),\left( 2,4 \right),\left( 3,3 \right)$ .
From the graph, we have to look for the points of the intersection of equations $y=2{{x}^{2}}$ and $y=6-x$ .
We can see that both the equations intersect at $\left( -2,8 \right)$ and $\left( 1.5,4.5 \right)$ (shown as red colour in the graph). The solution of $y=6-x$ will be the set of x coordinates of the points of intersection.
Hence, the solution of $2{{x}^{2}}+x-6=0$ is $\left\{ -2,1.5 \right\}$ .

Note: Students must draw the graph of $2{{x}^{2}}+x-6=0$ in the graph of $y=2{{x}^{2}}$ . They must note that the solution set will be the set of x coordinates of the points of intersection not the y-coordinates. They must know to simplify an equation. Students must make the second equation in terms of y.