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3 years ago

The quadratic equation y=2x^2-px+6 has equal roots. Find the possible values of p.

1 answer:

5 0

$\Delta=0\\
\Delta=(-p)^2-4\cdot2\cdot6=p^2-48\\
p^2-48=0\\
p^2=48\\
p=\sqrt{48} \vee p=-\sqrt{48}\\
p=4\sqrt3 \vee p=-4\sqrt3
$

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so 9n + 7 = 13 + 3n + 8n

then combine on other side so 9n + 7 = 13 + 11n

then subtract 7 to 13 so 6

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then subtract 9 to the other side so 2n

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In a quadratic equation

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Answer:

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Step-by-step explanation:

I've attached the graphs here:

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x=4

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