Question

Draw the graph of $ y = {x^2} + 2x - 3 $ and hence find the sum of roots of $ {x^2} - x - 6 = 0 $

Answer 1

Hint: In this question, we need to draw the graph of $ y = {x^2} + 2x - 3 $ and also, we need to determine the sum of roots of $ {x^2} - x - 6 = 0 $ . Here, we will determine the points of the equation $ y = {x^2} + 2x - 3 $ and plot the points in the graph. Then, using the formula of the sum of roots of a quadratic equation, we will determine the value of sum of roots.

Complete step-by-step answer:
Now, we will determine the points for the equation $ y = {x^2} + 2x - 3 $ ,
Let us consider $ x = 0 $ , then we have,
 $ y = {\left( 0 \right)^2} + 2\left( 0 \right) - 3 $
 $ y = - 3 $
Now, we got a points for the equation $ y = {x^2} + 2x - 3 $ , i.e., $ \left( {0, - 3} \right) $
Similarly, let us consider $ x = 1 $ , then we have,
 $ y = {\left( 1 \right)^2} + 2\left( 1 \right) - 3 $
 $ y = 1 + 2 - 3 $
 $ y = 0 $
Therefore, the point is $ \left( {1,0} \right) $ .
Let $ x = - 1 $ , we have,
 $ y = {\left( { - 1} \right)^2} + 2\left( { - 1} \right) - 3 $
 $ y = 1 - 2 - 3 $
 $ y = - 4 $
Therefore, the point is $ \left( { - 1, - 4} \right) $
Let $ x = 2 $ , we have,
 $ y = {\left( 2 \right)^2} + 2\left( 2 \right) - 3 $
 $ y = 4 + 4 - 3 $
 $ y = 5 $
Therefore, the point is $ \left( {2,5} \right) $
Let $ x = - 2 $ , we have,
  $ y = {\left( { - 2} \right)^2} + 2\left( { - 2} \right) - 3 $
 $ y = 4 - 4 - 3 $
 $ y = - 3 $
Therefore, the point is $ \left( { - 2, - 3} \right) $
Let $ x = 3 $ , we have,
 $ y = {\left( 3 \right)^2} + 2\left( 3 \right) - 3 $
 $ y = 9 - 6 - 3 $
 $ y = 12 $
Therefore, the point is $ \left( {3,12} \right) $
Let $ x = - 3 $ , we have,
 $ y = {\left( { - 3} \right)^2} + 2\left( { - 3} \right) - 3 $
 $ y = 9 - 6 - 3 $
 $ y = 0 $
Therefore, the point is $ \left( { - 3,0} \right) $

x	1	-1	2	-2	3	-3
y	0	-4	5	-3	12	0
P(x,y)	(1,0)	(-1,-4)	(2,5)	(-2,-3)	(3,12)	(-3,0)

Now let us plot it graphically,

Now, we need to determine the sum of the roots of the equation $ {x^2} - x - 6 = 0 $ .
We know that, for a quadratic equation $ a{x^2} + bx + c = 0 $ ,
The sum of the roots= $ \dfrac{{ - b}}{a} $
And, the product of the roots= $ \dfrac{c}{a} $
Therefore,
 $ {x^2} - \left( {sum\,of\,the\,roots} \right)x + \left( {product\,of\,the\,roots} \right) = 0 $
Therefore, by comparing $ {x^2} - x - 6 = 0 $ with the above equation, we have,
The sum of the equation= $ \dfrac{{ - \left( { - 1} \right)}}{1} $ $ = 1 $
Hence, the sum of roots of $ {x^2} - x - 6 = 0 $ is $ 1 $ .
So, the correct answer is “1”.

Note: It is important to note here that the sum of the roots of the quadratic equation is equal to the negation of the coefficient of the second term, divided by the leaving leading coefficient. The product of the roots of a quadratic equation is equal to the constant term, divided by the leading coefficient. In higher polynomials, the sum and the product of the roots vary. The roots of the quadratic equation can be given by, $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ .