First slide

Theory of equations

Question

$The roots α and β of the quadratic equation$ ${px}^{2} + qx + r = 0 are real and of opposite signs. The roots of$ $α (x - β)^{2} + β (x - α)^{2} = 0 are$

Moderate

A

positive
B

negetive
C

of opposite signs
D

non real

Solution

$\begin{array}{l} We have α + β = - \frac{q}{p}, α β = \frac{r}{p} \\ Now the given equation \\ \Rightarrow (α + β) x^{2} - 4 α β x + α β (α + β) = 0 \\ \Rightarrow (- \frac{q}{p}) x^{2} - 4 \frac{r}{p} x + \frac{r}{p} (- \frac{q}{p}) = 0 \\ \Rightarrow {pqx}^{2} + 4 prx + rq = 0 \\ Since α and β have opposite signs, therefore p and r \\ must have opposite signs. \\ \Rightarrow pq and rq must have opposite signs \Rightarrow {roots of equation pqx}^{2} + 4 prx + rq = 0 have opposite signs \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App