Draw the graph of given polynomial and find the zeroes p(x)= x^2-5x+6

Draw the graph of given polynomial and find the zeroes.
p(x)= x^2-5x+6

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[@khajaacademymathematica1729]

Thank you! Cheers!

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[@khajaacademymathematica1729]

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[@arhammohammed7948]

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[@khajaacademymathematica1729]

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[@khajaacademymathematica1729]

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[@mathsproblem3035]

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[@billanayak7540]

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[@khajaacademymathematica1729]

Ty

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100th comment

[@shanazsultanayasminsultana7954]

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[@keerthipriyanka2352]

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[@amanjunath247]

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[@khajaacademymathematica1729]

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Thank you for your support please share with your friends

[@kingsgaming4026]

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[@NoorjahanaNoorjahana-fd1zk]

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[@vamsibestha]

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[@vamsibestha]

1 reply

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[@khajaacademymathematica1729]

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[@samathanalla3340]

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[@user-pl4zp6vk7y]

Thank you so much sir I solve my graph in first attemp❤

[@khajaacademymathematica1729]

Thank you so much for your support.

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[@MOHAMMED-tb8yw]

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[@mukeshkumargurjar8972]

& There exist only one real zero of P(x) = (x - 2)(x ^ 2 + 3)
Reason; sum of zeroes of quadratic polynomials a * x ^ 2 + bx + c is (- b)/a
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the
correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true
sir answer please 🙏

[@khajaacademymathematica1729]

Option B
Amd sorry for the late reply

[@addebasultana1865]

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[@Sameer_652]

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[@khajaacademymathematica1729]

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[@harifacts007]

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[@ramyasurendharreddy7997]

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[@user-sj2qv6dk6p]

Thank you very much sir I have understood this process but I have a doubt from your perspective please can you clear it, it is that if the number is higher which it is not possible in the graph like 42, 30 can I leave that point like you leaved the first point 42 in exam question, mere marks tho nahi katenge na 😅

[@khajaacademymathematica1729]

Yes you can skip the higher values. Because your solution is where the parabola cuts x-axis which already you got. Thank you for your support

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Sirmalechapade😮

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[@khajaacademymathematica1729]

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[@MOHAMMED-tb8yw]

Good but some difficult

[@user-kr2wp1oi7u]

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[@balachandrabalarajupattigu1110]

TQ grapu

[@sultanbhai761]

Shukriya bhai

[@vghemanthsai6663]

Thank u sir