Question

12.
If p(x) = x2 + 3x - 2, then find the value of p(2) - P(-2) + p1/2
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Answer 1

Answer:

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Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{p(x) \: = {x}^{2} + 3x - 2 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{p(2) - p( - 2) + p(\dfrac{1}{2}) }  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

It is given

$\rm : \implies \:p(x) \: = \: {x}^{2} + 3x - 2$

Now, Consider p(2),

$\rm : \implies \:p(2) \: = \: {(2)}^{2} + 3 \times 2 - 2$

$\rm : \implies \:p(2) = 4 + 6 - 2$

$\rm : \implies \: \boxed{ \pink{ \rm \: p(2) = 8}}$

Now, Consider p( - 2)

$\rm : \implies \:p( - 2) \: = \: {( - 2)}^{2} + 3 \times ( - 2) - 2$

$\rm : \implies \:p( - 2) = 4 - 6 - 2$

$\rm : \implies \: \boxed{ \pink{ \rm \: p( - 2) = - 4}}$

Now, Consider p(1/2)

$\rm : \implies \:p(\dfrac{1}{2} ) \: = \: {(\dfrac{1}{2} )}^{2} + 3 \times \dfrac{1}{2} - 2$

$\rm : \implies \:p(\dfrac{1}{2} ) \: =\dfrac{1}{4} + \dfrac{3}{2} - 2$

$\rm : \implies \:p(\dfrac{1}{2} ) \: = \: \dfrac{1 + 6 - 8}{4}$

$\rm : \implies \: \boxed{ \pink{ \rm \: p(\dfrac{1}{2} ) = - \dfrac{1}{4} }}$

Hence, now Consider

$\rm : \implies \:p(2) - p( - 2) + p(\dfrac{1}{2})$

$\rm : \implies \:8 - ( - 4) - \dfrac{1}{4}$

$\rm : \implies \:12 - \dfrac{1}{4}$

$\rm : \implies \:\dfrac{48 - 1}{4}$

$\rm : \implies \:\dfrac{47}{4}$

Hence, the value of

$\rm : \implies \: \boxed{ \pink{ \rm \: p(2) - p( - 2) + p(\dfrac{1}{2}) = \dfrac{47}{4} }}$