Question

Let a, b be non-zero real numbers. Which of the following statements about the quadratic equation $ a{x^2} + \left( {a + b} \right)x + b = 0 $ is necessarily true?
i.It has at least one negative root.
ii.It has at least one positive root.
iiiBoth its roots are real.
A.(i) and (ii) only.
B.(ii) and (iii) only.
C.(ii) and (iii) only.
D.All of them

Answer 1

Hint: Here, we are given a quadratic equation and some statements about its roots. To find out which statement is correct, we need to find the roots of the given quadratic equation first. To find the roots of $ a{x^2} + \left( {a + b} \right)x + b = 0 $ , we will be using the quadratic formula.
 $ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $

Complete step-by-step answer:
In this question, we are given a quadratic equation $ a{x^2} + \left( {a + b} \right)x + b = 0 $ and we are given some statements about its roots and we need to find which statements are correct.
So, first of all let us find the roots of this equation.
To find the roots of a quadratic equation, we will use the quadratic equation. The quadratic formula is
 $ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Here, we have $ a = a $ , $ b = \left( {a + b} \right) $ and $ c = b $ .
Therefore, we get
 $ \Rightarrow x = \dfrac{{ - \left( {a + b} \right) \pm \sqrt {{{\left( {a + b} \right)}^2} - 4ab} }}{{2a}} $
Now, $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ . Therefore,
 $
\Rightarrow x = \dfrac{{ - \left( {a + b} \right) \pm \sqrt {\left( {{a^2} + 2ab + {b^2}} \right) - 4ab} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - \left( {a + b} \right) \pm \sqrt {{a^2} - 2ab + {b^2}} }}{{2a}} \;
  $
Now, we know that $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $ . Therefore,
 $
\Rightarrow x = \dfrac{{ - \left( {a + b} \right) \pm \sqrt {{{\left( {a - b} \right)}^2}} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - \left( {a + b} \right) \pm \left( {a - b} \right)}}{{2a}} \\
 $
 $
   \Rightarrow x = \dfrac{{ - a - b + a - b}}{{2a}} \\
   \Rightarrow x = \dfrac{{ - 2b}}{{2a}} \\
   \Rightarrow x = \dfrac{{ - b}}{a} \;
  $
And
 $
   \Rightarrow x = \dfrac{{ - a - b - a + b}}{{2a}} \\
   \Rightarrow x = \dfrac{{ - 2a}}{{2a}} \\
   \Rightarrow x = - 1 \;
  $
Therefore, the roots of the equation $ a{x^2} + \left( {a + b} \right)x + b = 0 $ are $ \dfrac{{ - b}}{a} $ and -1.
Now, let us see the given statements.
Statement 1: It has at least one negative root.
Here, both our roots are negative. Hence, statement 1 is correct.
Statement 2: It has at least one positive root.
Here, both our roots are negative. Hence, statement 2 is incorrect.
Statement 3: Both its roots are real.
Here, our roots are $ \dfrac{{ - b}}{a} $ and -1 which both are real. Hence, statement 3 is also correct.
Therefore, the correct answer is option B.
So, the correct answer is “Option B”.

Note: Following are the some important points about quadratic equations we must keep in mind.
Consider a quadratic equation: $ a{x^2} + bx + c = 0 $
I.Sum of roots $ = \dfrac{{ - b}}{a} $
II.Product of roots $ = \dfrac{c}{a} $