A quadratic equation is given by (a+b) x^2-(a+b+c) x+k=0 , where a, b, c are real. If k=(c)/(2),(c ≠ 0) , then the roots of the equation are:

Correct answer: B. Real and unequal If k= c 2 ,(c ≠ 0) then, the given equation is (a+b) x^2-(a+b+c) x+(c)/(2)=0 Now, D=\ -(a+b+c)\ ^2-4(a+b) · c 2 aligned & a^2+b^2+c^2+2 a b+2 b c+2 c a-2 a c-2 b c & (a+b)^2+c^2 > 0 ( since, c ≠ 0) aligned So, roots are real and unequal.

NDA 2023

Exam: NDA
Year: 2023
Subject: Mathematics
Chapter: Quadratic Equation

NDA 2023 · Mathematics · Quadratic Equation

A quadratic equation is given by (a+b) x^2-(a+b+c) x+k=0 , where a,...

A quadratic equation is given by

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

, where

a, b, c

are real. If

k=\frac{c}{2},(c \neq 0)

, then the roots of the equation are:

Options

AReal and equal
BReal and unequal
CReal iff a > c
DComplex but not real

Correct answer

B. Real and unequal

Step-by-step solution

\mathrm{k}=\frac{\mathrm{c}}{2},(\mathrm{c} \neq 0)

then, the given equation is

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

Now,

\quad \mathrm{D}=\{-(\mathrm{a}+\mathrm{b}+\mathrm{c})\}^2-4(\mathrm{a}+\mathrm{b}) \cdot \frac{\mathrm{c}}{2}

\begin{aligned} & a^2+b^2+c^2+2 a b+2 b c+2 c a-2 a c-2 b c \\ & (a+b)^2+c^2 > 0 \quad(\text{ since, } c \neq 0) \end{aligned}

$$ So, roots are real and unequal.

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