Question

prove that for any real value of a , (a²+3)x²+(a+2)x-5<0 is satisfied for atleast one negative x.​

Answer 1

Answer:

Step-by-step explanation:

(a²+3)x²+(a+2)x-5<0

according to Quadratic Formula,

x ₁ = $\frac{-(a+2) + \sqrt{(a+2)^{2} - 4(a^{2}+3 )(-5)} }{2(a^{2}+3)}$    , x₂ = $\frac{-(a+2) - \sqrt{(a+2)^{2} - 4(a^{2}+3 )(-5)} }{2(a^{2}+3)}$

x₁ =  $\frac{-(a+2) + \sqrt{(a+2)^{2} +20(a^{2}+3)} }{2(a^{2}+3)}$      , x₂= $\frac{-(a+2) - \sqrt{(a+2)^{2} +20(a^{2}+3)} }{2(a^{2}+3)}$

 

here, $\sqrt{(a+2)^{2} +20(a^{2}+3)}$  is a positive number for any value of a and greater than a+2  

$\sqrt{(a+2)^{2} +20(a^{2}+3)}$   >  a+2  >  - (a+2)

so ,  x₂ is negative for any real value of a

for any real value of a , (a²+3)x²+(a+2)x-5<0 is satisfied for atleast one negative x