If a 0, b 0, c 0, then both the roots of the equation a x2+b x+c=0

Question

If a> 0, b> 0, c> 0, then both the roots of the equation a x2+b x+c=0 (A) Are real and negative (B) Have negative real parts (C) Are rational numbers (D) None of these

Tardigrade Admin · Accepted Answer

The roots of the equations are given by

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

(i) Let

b^{2} - 4 a c > 0, b > 0

Now, if

a > 0, c > 0, b^{-} - 4 a c < b^{-}

\Rightarrow

the roots are negative.
(ii) Let

b^{2} - 4 a c < 0

, then the roots are given by

x = \frac{- b \pm i ( 4 a c - b ^{2} )}{2 a}, (i = - 1)

which are imaginary and have negative real part

[∵ b > 0]

∴

In each case, the roots have negative real part

Q. If a>0,b>0,c>0, then both the roots of the equation ax2+bx+c=0

Are real and negative

Have negative real parts

Are rational numbers

None of these

Q. If $a > 0, b > 0, c > 0$ , then both the roots of the equation $a x^{2} + b x + c = 0$