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Byju's Answer

Standard XII

Mathematics

Nature of Roots

The roots of ...

Question

The roots of the following quadratic equation Real and unequal.
$(b + c) x^{2}$ - (a + b + c) x + a = 0

True

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False

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Solution

The correct option is A True
$(b + c) x^{2} - (a + b + c) x + a = 0$
$\Rightarrow$ Here, $A = b + c, B = - (a + b + c), C = a$
$\Rightarrow$ $D = B^{2} - 4 A C$
$= [- (a + b + c)]^{2} - 4 (b + c) (a)$
$= a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a - 4 b c - 4 c a$
$= a^{2} + b^{2} + c^{2} + 2 a b - 2 b c - 2 c a$
$= (a + b - c)^{2}$
$∴$ $D > 0$
$∴$ We can see discriminant is greater than $0$ , so nature of roots are real and unequal.

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A, G

and

H

are respectively arithmetic, geometric and harmonic means between

a

and

b

both being unequal and positive, then

A = \frac{a + b}{2} \Rightarrow a + b = 2 A, G = \sqrt{a b} \Rightarrow a b = G^{2}

and

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From the above discussion we can say that

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The roots of the following quadratic equation Real and unequal.(b+c)x2 - (a + b + c) x + a = 0

The correct option is A True(b+c)x2−(a+b+c)x+a=0⇒ Here, A=b+c,B=−(a+b+c),C=a⇒ D=B2−4AC =[−(a+b+c)]2−4(b+c)(a) =a2+b2+c2+2ab+2bc+2ca−4bc−4ca =a2+b2+c2+2ab−2bc−2ca =(a+b−c)2∴ D>0∴ We can see discriminant is greater than 0, so nature of roots are real and unequal.

The roots of the following quadratic equation Real and unequal.
$(b + c) x^{2}$ - (a + b + c) x + a = 0