Question

A quadratic polynomial, whose zeros are -3 and 4
a. $ {x^2} - x + 12 $
b. $ {x^2} + x + 12 $
c. $ \dfrac{{{x^2}}}{2} - \dfrac{x}{2} - 6 $
d. $ 2{x^2} + 2x - 24 $

Answer 1

Hint: To solve the question there is no direct method. We solve each and every equation in the options and find their roots and we are going to compare the obtained roots and the roots mentioned in the question.

Complete step-by-step answer:
We find the roots of each and every equation which are given in the options.
Now consider the first option the equation is given as $ {x^2} - x + 12 $
To find the roots we use the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Here a = 1 b= -1 and c = 12. substituting all the values in the formula we have
 $
   \Rightarrow x = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(1)(12)} }}{{2(1)}} \\
   \Rightarrow x = \dfrac{{1 \pm \sqrt {1 - 48} }}{2} \\
   \Rightarrow x = \dfrac{{1 \pm 7i}}{2} \;
  $
Here we get the roots are imaginary. Therefore, the option is not the correct one.
Now consider the second option the equation is given as $ {x^2} + x + 12 $
To find the roots we use the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Here a = 1 b= 1 and c = 12. substituting all the values in the formula we have
 $
   \Rightarrow x = \dfrac{{ - (1) \pm \sqrt {{{(1)}^2} - 4(1)(12)} }}{{2(1)}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - 48} }}{2} \\
   \Rightarrow x = \dfrac{{1 \pm 7i}}{2} \;
  $
Here we get the roots are imaginary. Therefore, option b is not the correct one.
Now consider the third option the equation is given as $ \dfrac{{{x^2}}}{2} - \dfrac{x}{2} - 6 $
To find the roots we use the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Here $ a = \dfrac{1}{2} $ $ b = - \dfrac{1}{2} $ and c = -6. substituting all the values in the formula we have
 $
   \Rightarrow x = \dfrac{{ - \left( { - \dfrac{1}{2}} \right) \pm \sqrt {{{\left( { - \dfrac{1}{2}} \right)}^2} - 4\left( {\dfrac{1}{2}} \right)( - 6)} }}{{2\left( {\dfrac{1}{2}} \right)}} \\
   \Rightarrow x = \dfrac{{\dfrac{1}{2} \pm \sqrt {\dfrac{1}{4} + 12} }}{1} \\
   \Rightarrow x = \dfrac{1}{2} \pm \sqrt {\dfrac{{49}}{4}} \\
   \Rightarrow x = \dfrac{1}{2} \pm \dfrac{7}{2} = \dfrac{{1 \pm 7}}{2} \;
  $
 $ \Rightarrow x = \dfrac{{1 + 7}}{2} $ and $ \Rightarrow x = \dfrac{{1 - 7}}{2} $
 $ \Rightarrow x = 4 $ and $ x = - 3 $
Here we get the roots and they are matching to the roots which is in the question Therefore the option c is the correct one.
We got the equation and whose roots are satisfying the roots which are mentioned in the question. Therefore the $ \dfrac{{{x^2}}}{2} - \dfrac{x}{2} - 6 $ is a quadratic polynomial whose zeros are -3 and 4.
So, the correct answer is “Option C”.

Note: By factorising we can find the roots and we can find the solution. By using formulas, we are able to find the roots in the correct manner. Here we need to solve the equation to select the correct option. Once our answer matches the question and there is no need to move forward to solve another equation.