Theory of EquationsmediumFree

Theory of Equations: Non Zero Real Numbers Roots Roots Absolute Value

JEE Maths question with a full step-by-step solution.

Question

a, b, c, d

are non-zero real numbers such that

c

and

d

are the roots of

x^2 + ax + b = 0

and

a

and

b

are the roots of

x^2 + cx + d = 0

, then the absolute value of

a + 2b + 3c + 4d

2

4

6

8

correct

Solution

Step 1: From the two equations:

c + d = -a, \quad cd = b \quad \cdots (1), \qquad a + b = -c, \quad ab = d \quad \cdots (2)

Step 2: From

c+d=-a

and

a+b=-c

: adding gives

a+b+c+d = -(a+c)

, so

b+d = -2(a+c)

. From

c+d=-a

d-b=0-b-c-...

Let me use direct algebra. From (1) and (2):

c+d=-a

... (I) and

a+b=-c

... (II). Subtracting:

c+d-a-b = -a+c \implies d-b=0 \implies d = b

. Step 3:

d = b \implies ab = d = b \implies b(a-1) = 0

. Since

b \neq 0

a = 1

. Step 4: From

a + b = -c

1+b = -c

, so

c = -1-b

. From

cd = b

(-1-b)b = b \implies -1-b = 1 \implies b = -2

(since

b \neq 0

). So

c = 1

d = -2

. Step 5:

|a + 2b + 3c + 4d| = |1 + 2(-2) + 3(1) + 4(-2)| = |1-4+3-8| = |-8| = 8

. Correct answer: (4)

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