GATE Overflow Test Series | Quantitative Aptitude | Test 1 | Question: 23

Question

If the difference between the roots of the equation $x^{2} + ax + 2 = 0$ is less then $\sqrt{56},$ then the set of possible values of $a$ is ________

• A. $(-8,8)$
• B. $(4,\infty)$
• C. $(-\infty, 8)$
• D. $[-4,4]$

Answer 1

Lets say roots of the quadratic equation is $\alpha, \beta.$
  
  From the quadratic equation $x^{2} + ax + 2 = 0$
  
  Sum of roots $\alpha + \beta = \dfrac{-a}{1} = -a$
  
  Product of roots $\alpha\beta = \dfrac{2}{1} = 2$
  
  We know that $(\alpha - \beta)^{2} = \alpha^{2} + \beta^{2} - 2\alpha \beta$
  
  $\implies (\alpha - \beta)^{2} = \alpha^{2} + \beta^{2} + 2\alpha \beta -  4\alpha \beta $
  
  $\implies (\alpha - \beta)^{2} =  (\alpha + \beta)^{2}-  4\alpha \beta $
  
  $\implies \mid \alpha - \beta \mid = \sqrt{(\alpha + \beta)^{2}-  4\alpha \beta }$
  
  $\implies  \mid \alpha - \beta \mid = \sqrt{(-a)^{2} - 8}$
  
  $\implies  \mid \alpha - \beta \mid = \sqrt{a^{2} - 8}$
  
  According to the question, $\mid \alpha - \beta \mid < \sqrt{56}$
   
  $\implies \sqrt{(a^{2} - 8)}<\sqrt{56}$
  
  $\implies a^{2} - 8 < 56$
   
  $\implies a^{2} < 64$
  
  $\implies -8 < a < 8$
  
  $\implies a\in(-8,8).$
  
  So, the correct answer is $(A).$

Comments

Given

$a^{2} < 64 \\ a < 8 \hspace{1cm} or \hspace{1cm} a<-8$

hence couldn’t $a$ take values less than $-8$ till $-\infty$, meaning option C?

Please correct me if I’m wrong.

---

Here ,

⟹sqrt(a^2 – 8 ) < sqrt ( 56)

so acc. to this condition  {-2,-1,0,1,2} are not satisfying this as sqrt of -ve is imaginary ..so aren’t these point excluded from the interval or this range[-2,2] should be excluded?

Please correct me If I’m wrong.

---

you forgot to add one more constraint which is D should be >= 0..
so ans will (-8, -2root) Union (2root2, 8)