Question

If the equations x² + ax + 12 = 0, x² + bx + 15 = 0 and x² + (a + b)x + 36 = 0 have a common positive root, then ab equals to ________.​

Answer 1

Answer:

answer 3

x2 +ax +12=0,_(1)

x2+bx+15=0,_(2)

x2+(a+b )x+36=0_(3)

let the common root be m

so,m will setisty all three equation

m2

$m^{2} + am + 12 = 0| \: (4) \: \\ m{2} + 6m + 15 = 0 \: (5) \: \\ m ^{2} + (a + b)m + 36 = 0(6) \:$

ading 5and 6

2am का स्क्वायर प्लस ए ब्रैकेट ए प्लस बी एम प्लस 7 इक्वल टू जीरो subtraining 6from7

एम का स्क्वायर माइनस 9 स्क्वायर एमएम 1 लाइन और इक्वल - 3 पैकेट व्हिच इज नॉट पॉसिबल

Answer 2

Solution!!

x² + ax + 12 = 0...(1)

x² + bx + 15 = 0...(2)

x² + (a + b)x + 36 = 0...(3)

Adding (1) and (2), we get,

2x² + (a + b)x + 27 = 0...(4)

Subtracting (3) and (4), we get,

-x² + 9 = 0

x² = 9

x = √9

x = 3

Putting the value of 'x' in (1),

x² + ax + 12 = 0

3² + 3a + 12 = 0

9 + 3a + 12 = 0

3a + 21 = 0

3a = -21

a = -7

Putting the value of 'x' in (2),

x² + bx + 15 = 0

3² + 3b + 15 = 0

9 + 3b + 15 = 0

3b + 24 = 0

3b = -24

b = -8

ab = -7 × (-8)

ab = 56

If the equations x² + ax + 12 = 0, x² + bx + 15 = 0 and x² + (a + b)x + 36 = 0 have a common positive root, then ab equals to 56.