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@Syed Hamza Khalid, as you said there is no possibility for $p,q$ to be integers. So, the question must be having a typo or it wants to say rational numbers instead of integers.

Even I don't know- I got $-2( x + \dfrac{3}{2} )^2 - \dfrac{21}{2}$ but the question asked $a, p$ and $q$ are integers.. and these are not $\therefore$ I think there is a typo error in the question perhaps. But how did u get $+ \dfrac{39}{2}$ at the end

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## Comments

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TopNewestHere is my full solution :

$\begin{aligned} a(x + p)^2 + q & = - 2x^2 - 6x + 15 \\ a(x^2 + 2px + p^2) + q & = -2x^2 - 6x + 15 \\ {\color{#3D99F6}a}x^2 + {\color{#E81990}2ap}x + {\color{#20A900} ap^2 + q} & = {\color{#3D99F6}-2}x^2 {\color{#E81990}-6}x + {\color{#20A900}15} \\ \end{aligned}$

On comparing both the sides we get, $a = -2, p = \dfrac32, q = \dfrac{39}{2}$. So, the final result is

$-2x^2 - 6x + 15 = -2 \left( x - \dfrac32 \right)^2 + \dfrac{39}{2}$

@Syed Hamza Khalid, as you said there is no possibility for $p,q$ to be integers. So, the question must be having a typo or it wants to say rational numbers instead of integers.

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Yeah thats wat I think

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Even I don't know- I got $-2( x + \dfrac{3}{2} )^2 - \dfrac{21}{2}$ but the question asked $a, p$ and $q$ are integers.. and these are not $\therefore$ I think there is a typo error in the question perhaps. But how did u get $+ \dfrac{39}{2}$ at the end

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$-ap^2 + q = 15$ from which we will get q.

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I just used complete the square: $= -2x^2 - 6x +15 \\ = -2(x^2 + 3x) + 15 \\ = -2((x + \dfrac{3}{2})^2 - \dfrac{9}{4} ) - 15 \\ = -2(x + \dfrac{3}{2})^2 + \dfrac{9}{2} - 15 \\ = -2(x + \dfrac{3}{2} )^2 - \dfrac{21}{2} . \square$

This was the method we learned at school $\therefore$ I applied it but I don't understand why I didn't get integers

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I got it as $-2(x + \dfrac32)^2 + \dfrac{39}{2}$. Is it correct.

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Hah, what a question! Have you used a research paper service like https://ca.edubirdie.com/research-paper-writing-services to complete these tasks or not? I'm gonna pass my exams soon and want to know the answer.

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