Mathematics: Prove algebraically that (m + 2)2 – m 2 – 12 is always a multiple of 4

Center of circle is ( 1, 0).Slope of normal passing though ( 1, 0 ) and ( 2, -5 ) is :So, slope of tangent will be :Equation of tangent :Hence, this is the required solution....

Prove algebraically that (m + 2)2 – m 2 – 12 is always

Ответ:

ntrann22

20.09.2021

Mathematics

$(m + 2)^{2} - m^{2} - 12 = 4(m - 2)$

Step-by-step explanation:

Step 1:

Write the expression

$(m+2)^{2} - m^{2} - 12$

Step 2: Expand $(m + 2)^{2}$

$(m+2)^{2} - m^{2} - 12\\(m+2)(m+2) - m^{2} - 12\\m^{2} + 2m + 2m + 4 - m^{2} - 12$

Step 3: Collect similar terms

$m^{2} - m^{2} + 4m + 4 - 12\\4m - 8$

Step 4: Factor 4 out of the expression to prove that the expression is a multiple of 4.

$Therefore\\4m - 8 = 4(m - 2)\\Hence,\\(m+2)^{2} - m^{2} - 12 is a multiply of 4 because the expression is equal to 4(m-2)$

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Ответ:

hncriciacheichi

25.03.2022

Mathematics

Center of circle is ( 1, 0).

Slope of normal passing though ( 1, 0 ) and ( 2, -5 ) is :

$m_p=\dfrac{0-(-5)}{1-2}\\\\m_p=-5$

So, slope of tangent will be :

$m_t=-\dfrac{1}{m_n}\\\\m_t=\dfrac{1}{5}$

Equation of tangent :

$y-(-5)=\dfrac{1}{5}(x-2)\\\\5y+25=x-2\\\\5y-x+27=0$

Hence, this is the required solution.

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Prove algebraically that (m + 2)2 – m 2 – 12 is always a multiple of 4

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