You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If r be the...

Question

If $r$ be the ratio of the roots of the equation $a x^{2} + b x + c = 0$ , then $\frac{{(r + 1)}^{2}}{r} =$

\frac{a^{2}}{b c}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{b^{2}}{c a}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{c^{2}}{a b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{b^{2}}{c a}$
Given equation is $a x^{2} + b x + c = 0$ ...(1)
Let the root of the equation (1) be $α$ and $r α$ , then
$α + r α = - \frac{b}{a}$ ...(2)
and $r α^{2} = \frac{c}{a}$ ...(3)
From (2) $α = - \frac{b}{a (r + 1)}$
Putting the value of $α$ in (3), we get
$\frac{r b^{2}}{a^{2} {(r + 1)}^{2}} = \frac{c}{a} \Rightarrow \frac{b^{2}}{a c} = \frac{{(r + 1)}^{2}}{r}$

Suggest Corrections

Similar questions

Q. If the quadratic equation

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

, where

a, b, c

are distinct real numbers, has imaginary roots, then

Q. If

r

be the ratio of the roots of the equation

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

, show that

r b^{2} = {(r + 1)}^{2} a c

Q. Prove that

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) \geq 0

for all

x \in R .

If equality holds then find the ratio of the roots of the equation

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

Consider the quadratic equation

a x^{2} + b x + (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 0

(where

a, b, c

are distinct real numbers)

If the equation does not have any real roots then which of the following is always true.

Q. Lets consider quadratic equation

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

where

a, b, c \in R

and

a \neq 0

. If above equation has roots

α, β

, then

α + β = - \frac{b}{a}, α β = \frac{c}{a}

and the equation can be written as

a x^{2} + b x + c = a (x - α) (x - β)

. Also, if

a_{1}, a_{2}, a_{3}, a_{4}

, ..... are in A.P., then

a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = . . . \neq 0

and if

b_{1}, b_{2}, b_{3}, b_{4}

, ... are in G.P., then

\frac{b_{2}}{b_{1}} = \frac{b_{3}}{b_{2}} = \frac{b_{4}}{b_{3}} =

...

\neq 1

Now if

c_{1}

c_{2}

c_{3}

c_{4}

, ... are in HP, then

\frac{1}{c_{2}} - \frac{1}{c_{1}} = \frac{1}{c_{3}} - \frac{1}{c_{2}} = \frac{1}{c_{4}} - \frac{1}{c_{3}} =

...

\neq 0

. If the roots of equation

a (b - c) x^{2} + b (c - a) x + c (a - b) = 0

are equal, then

a, b, c

are in

Join BYJU'S Learning Program

If r be the ratio of the roots of the equation ax2+bx+c=0, then (r+1)2r=

The correct option is B b2caGiven equation is ax2+bx+c=0 ...(1)Let the root of the equation (1) be α and rα, then α+rα=−ba ...(2)and rα2=ca ...(3)From (2) α=−ba(r+1)Putting the value of α in (3), we getrb2a2(r+1)2=ca⇒b2ac=(r+1)2r

If $r$ be the ratio of the roots of the equation $a x^{2} + b x + c = 0$ , then $\frac{{(r + 1)}^{2}}{r} =$