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anonymous

  • 4 years ago

Just another problem: A quadratic polynomial \( P(x) \) is such that \( P(x) \) never takes any negative values and \( P(0)=8 \) and the \( P(8)=0\). Find \( P(-4) \) Genre: algebra pre-calculus Rating: Easy

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  1. Zarkon
    • 4 years ago
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    18

  2. anonymous
    • 4 years ago
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    @zarkon: could you please explain with steps

  3. anonymous
    • 4 years ago
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    It is just too easy for Zarkon :)

  4. Zarkon
    • 4 years ago
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    you did clasify this as easy :)

  5. Mr.Math
    • 4 years ago
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    Let \(P(x)=ax^2+bx+c\), then \(P(0)=8 \implies c=8\) and \(P(8)=0 \implies 8^2a+8b+8=0 \implies b=-8a-1\). Now, \(P(x)=ax^2-(8a+1)x+8\ge 0 \implies (x-8)(ax-1)\ge 0 \implies a=\frac{1}{8}.\) Thus \(P(-4)=\frac{1}{8}(-4)^2-2(-4)+8=2+8+8=18.\)

  6. Zarkon
    • 4 years ago
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    i used the fact that (8,0) must be the vertex to get the 2nd equation

  7. Zarkon
    • 4 years ago
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    used the vertex formula

  8. Mr.Math
    • 4 years ago
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    Oh that's better I think.

  9. ash2326
    • 4 years ago
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    Let the quadratic polynomial be \[P(x)=ax^2+bx+c\] as it's given it doesn't take negative values D<0 \[b^2-4ac<0\] let's substitute x=0 P(0)=8 c=8 P(8)=64a+8b+8=0 or \[8a+b+1=0\] we have b^2-4ac<0 or b^2-32a<0 or b^2<32 a we have 8a=-1-b b^2<-4-4b b^2+4b+4<0 (b+2)^<0 so b is a complex no. of the form =-2-xi now let x be 1 so \[P(x)=\frac{-(1+i)}{8} x^2+(-2+i)x+8\] Could anyone point out my mistake??

  10. Zarkon
    • 4 years ago
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    you way is good

  11. anonymous
    • 4 years ago
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    Seems I am only one who used Maxima-minima.

  12. anonymous
    • 4 years ago
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    maxima-minima : u mean by taking the second derivatives of the eqs and substituting as rt-s^2 ??

  13. Mr.Math
    • 4 years ago
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    We can also use the derivative (I don't know if that what you mean @Fool). As Zarkon said, the polynomial must have its minimum at x=8. So \(P′(x)=2ax+b=0 \text{ at } x=8.\)

  14. Mr.Math
    • 4 years ago
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    i.e. \(16a+b=0\).

  15. anonymous
    • 4 years ago
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    Yes Mr.Math you got that right :)

  16. Mr.Math
    • 4 years ago
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    :-)

  17. phi
    • 4 years ago
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    @ash if the discriminant is less than zero the square root is imaginary and there are ZERO x-intercepts however,the problem states P(8)= 0. so we must have a repeated root at x=8, and the discriminant = 0 Zarkon's approach seems to fastest: given 8 is a root, the equation is y= a(x-8)^2 + b 0= a(8-8)^2 + b --> b=0 8= a(-8)^2 --> a= 1/8 y= (1/8)*(-4-12)^2 = 144/8 = 18

  18. ash2326
    • 4 years ago
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    Thanks Phi, got it :)

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