find the value of p , if the equarions 3x square -2x +p=0 and 6x square -17x +12=0 have common root
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Answered by
44
first we have to find the zeroes of 2nd equation
6x^2 - 17x +12 = 0
6x^2 -9x -8x + 12 = 0
3x(2x - 3) -4(2x - 3)
(3x - 4) (2x - 3)
x = 4/3 , x = 3/2
now we first find relation between zeroes
alpha × beeta = c/a
4/3 × 3/2 = 12/6
2 = 2
now in first equation
3x^2 - 2x + p = 0
alpha × beeta = c/a
2 = p/ 3
p = 6
6x^2 - 17x +12 = 0
6x^2 -9x -8x + 12 = 0
3x(2x - 3) -4(2x - 3)
(3x - 4) (2x - 3)
x = 4/3 , x = 3/2
now we first find relation between zeroes
alpha × beeta = c/a
4/3 × 3/2 = 12/6
2 = 2
now in first equation
3x^2 - 2x + p = 0
alpha × beeta = c/a
2 = p/ 3
p = 6
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Answered by
11
For equation ,
6x²-17x+12=0
6x²-9x-8x+12=0
3x(2x-3)-4(2x-3)=0
(2x-3)(3x-4)=0
2x-3=0 or 3x-4=0
x=3/2 or x=4/3
x=3/2 , 4/3
Since , both the equation have common root
So x=3/2 , 4/3 will also be the roots of equation 3x²-2x+p=0
Substituting x=3/2
3(3/2)²-2×(3/2)+p=0
3×9/4-3+p=0
27/4-3+p=0
27-12/4+p=0
15/4+p=0
p=-15/4
Substituting x=4/3
3×(4/3)²-2(4/3)+p=0
3×16/9-8/3+p=0
16/3-8/3+p=0
16-8/3+p=0
8/3+p=0
p=-8/3
Therefore , the value of p is (-15/4 , -8/3)
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