Math, asked by RanjaniR, 11 months ago

Find the zero's of
| P(x)= x²-7x +12​

Answers

Answered by varun164323
5

Step-by-step explanation:

x ^2 -7x+12

x^2-4x-3x+12

x(x-4) 3(x-4)

(x-4) (x+3)

now x-4=0

x= 4

in second bracket x+3= 0

x= -3

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Answered by BrainlyConqueror0901
88

Answer:

\huge{\boxed{\boxed{\sf{x=3\:and\:4}}}}

Step-by-step explanation:

\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}

P(x)= x²-7x +12

To find zeroes of this eqn:

\huge{\boxed{\boxed{\underline{\sf{METHOD(1)-}}}}}

Middle term spliting:

 {x}^{2}  - 7 x + 12  = 0\\  = ) {x}^{2}  - 4x - 3x + 12 = 0 \\  = )x(x - 4) - 3(x - 4) = 0 \\  = )(x - 3)(x - 4) = 0 \\  = )x - 3 = 0 \\  = )x = 3 -  -  -  -  - 1st  \: zeroes \\   = )x - 4 = 0 \\  = )x = 4 -  -  -  -  - 2nd \: zeroes

\huge{\boxed{\boxed{\underline{\sf{METHOD(2)-}}}}}

Quadratic formula:

 {x}^{2}  - 7x + 12 \\ d =  {b}^{2}  - 4ac \\ d =  {7}^{2}  - 4(1 \times 12) \\ d = 49 - 48 \\ d = 1 \\ x =  \frac{ - b +  \sqrt{d} }{2a}  \\ x =  \frac{  - ( - 7) +   \sqrt{1} }{2}  \\ x =  \frac{7 + 1}{2}  \\x  =  \frac{8}{2}  \\ x = 4  -  -  -  -  - 1st \: zeroes\\ similarly \\  x = \frac{7 - 1}{2}  \\ x =  \frac{6}{2}  \\ x = 3 -  -  -  -  - 2nd \: zeroes

\huge{\boxed{\boxed{\sf{x=3\:and\:4}}}}

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