(a) If the roots $α, β$ of $a x^{2} + b x + c = 0$ be real, then establish between the coefficients under the following conditions:
(i) Roots are equal and opposite.
(ii) Roots are of opposite signs.
(iii) Roots are both $- i v e$
(iv) Roots are both $+ i v e$ .
(b) Let $a > 0, b > 0$ , then both roots of the equation $a x^{2} + b x + c = 0$
(i) are real and negative
(ii) have negative real parts.
(iii) none of these.
(c) If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary. then for all real values of x, the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is
(a) less than $- 4 a b$
(b) greater then $4 a b$
(c) less then $4 a b$
(d) greater than $- 4 a b$

Open in App

Solution

(a) We have $Δ = b^{2} - 4 a c \geq 0, S = - \frac{b}{a}, P = \frac{c}{a}$
(i) $S = 0 \Rightarrow b = 0$ , $Δ \geq 0 \Rightarrow a c = - i v e$
$∴$ a andc are of opposite signs.
(ii) $P = - i v e$ $∴$ $\frac{c}{a} = - i v e$
i.e. a and c are of opposite signs so that ac is $- i v e$ and $Δ = b^{2} - 4 a c \geq 0$
(iii) $S = - i v e, P = + i v e$
$- \frac{b}{a} = - i v e \Rightarrow \frac{b}{a} = + i v e$
$∴$ a,b are of same sign either $+ i v e$ or $- i v e$ . In this case $P = \frac{c}{a} = + i v e$ i.e. both c and a have the sign either $+ i v e$ or $- i v e$ .
$\Rightarrow$ a,b and a,c have the same sign.
(iv) $S = + i v e, P = + i v e$
$- \frac{b}{a} = + i v e$ $∴ α, b$ of opposite sign.......(1)
$\frac{c}{a} = + i v e$ $∴$ c and a have same sign.......(2)
If a,c are $+ i v e$ , then b is $- i v e$ by (1)
If a,c are $- i v e$ , then b is $+ i v e$ by (1)
$∴$ a,c are of same sign and b si of opposite sign.
(b) Ans. (i), (ii).
Roots are given by $x = \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ .......(1)
since $a > 0, b > 0$ , we have $- \frac{b}{2 a} < 0$ . If $b^{2} - 4 a c < 0$ , then roots are imaginary of which the real part $- \frac{b}{2 a}$ is negative. If $b^{2} - 4 a c > 0$ then the roots are real. If $a c = + i v e$ i.e. c is $+ i v e$ as a is given to be $+ i v e$ then in this case $\sqrt{b^{2} - 4 a c} < b$ . Hence both the roots will be real and $- i v e$ from (1)
If however, $a c = - i v e$ i.e. c is $- i v e$ then $\sqrt (b^{2} - 4 a c) > b$ then one root will be $+ i v e$ and other $- i v e$
(c) Ans.(d).
Given $c^{2} - 4 a b < 0$ , i.e., $c^{2} < 4 a b$ ........(1)
Given expression is
$3 {(b x + c)}^{2} - c^{2} \geq - c^{2} > - 4 a b$ by (1)

Suggest Corrections

Similar questions

If the roots of the equation $b x^{2} + c x + a = 0$ be nonreal, then for all real values of x, the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

a + b + c > 0

a < 0 < b < c

a (x - b) (x - c) + b (x - c) (x - a) + c (x - a) (x - b) = 0

A x^{2} + B x + K = 0

B^{2} - 4 A K > 0

a x^{2} + b x + c = 0

x^{2} + b x + c = 0

1

x = b + c + 1

s

α, β

a x^{2} + b x + c = 0, (b \neq 0)

α^{4}, β^{4}

l x^{2} + m x + n = 0

α, β

a^{2} l x^{2} - 4 a c l x + 2 c^{2} l + a^{2} m = 0

Join BYJU'S Learning Program