Question

3. The sum of the non-real roots of (x² + x-2)(x2 + x-3) = 12 is (1) -1 (2) 1 (3) -6 (4) 6​

Answer 1

Answer:

The value is sum of non real values =-1

Step-by-step explanation:

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Answer 2

The sum of the non-real roots of (x² + x-2)(x2 + x-3) = 12 is (1) -1 (2) 1 (3) -6 (4) 6The sum of the non-real roots of (x² + x-2)(x2 + x-3) = 12 is (1) -1 (2) 1 (3) -6 (4) 6We have,

(x

2

+x−2)(x

2

+x−3)=12

Let, (x

2

+x−2)=y

Then,

y(y−1)=12

y

2

−y=12

y

2

−y−12=0

y

2

−(4−3)y−12=0

y

2

−4y+3y−12=0

y(y−4)+3(y−4)=0

(y−4)(y+3)=0

If,y−4=0 then,y=4

If,y+3=0 then,y=−3

Put the value of y=(x

2

+x−2)

(x

2

+x−2)=4

x

2

+x−6=0

x

2

+(3−2)x−6=0

x

2

+3x−2x−6=0

x(x+3)−2(x+3)=0

(x+3)(x−2)=0

x=2,−3

If, y=−3

x

2

+x−2=−3

x

2

+x−1=0

Then, sum of non real values=−1

Hence, this is the answer.

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