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Mind sharing your solution bro? I'm kinda stuck with this
x²−6x+9+16 + x²+10x+25+144
=(x−3)²+4² + (x+5)²+12²
I know solving the x-intercept of the line passing through the points (3,−4) and (−5,12) or (3,4) and (−5,−12) will give the answer but I don't know why is it like that.
@Vincent Lopez
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You can find the derivative of the function and set that equal to 0. Then solve for x. To make sure that it is a minimum point, take the second derivative of the function to see whether it is concave up on the stationary point (concave up means that dx2d2y>0 )
@James Watson
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When I thought I was done but completely forgotten the most important part, my eyes can't keep up anymore. I'll check it tomorrow bro, thanks for helping me again :)
@Vincent Lopez
–
If the y coordinate is 4, the distance from P to A won't change. However, P cannot be on segment AB, so the triangle inequality's equality won't hold.(i.e. Choose A and B such that A and B are on different side of the X axis
Easy Math Editor
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x=1
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Mind sharing your solution bro? I'm kinda stuck with this
x²−6x+9+16 + x²+10x+25+144
=(x−3)²+4² + (x+5)²+12²
I know solving the x-intercept of the line passing through the points (3,−4) and (−5,12) or (3,4) and (−5,−12) will give the answer but I don't know why is it like that.
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i used calculus for this; idk the algebraic way. If you want to know the calculus route i can share it
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x. To make sure that it is a minimum point, take the second derivative of the function to see whether it is concave up on the stationary point (concave up means that dx2d2y>0 )
You can find the derivative of the function and set that equal to 0. Then solve forLog in to reply
f(x)=x²−6x+25+x²+10x+169
f′(x)=x²−6x+25x−3+x²+10x+169x+5=0
x²+10x+169x+5=−x²−6x+25x−3
Squaring both sides of the equation
x²+10x+169(x+5)²=x²−6x+25(3−x)²
The reciprocals of equals are equal
(x+5)²x²+10x+169=(3−x)²x²−6x+25
(x+5)²(x+5)²+12²=(3−x)²(x−3)²+4²
1+(x+5)²12²=1+(3−x)²4²
(x+5)²12²=(3−x)²4²
Get the square root of both sides
x+512=3−x4
x=1
Is this the right way bro?
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f(x)=x²−6x+25+x²+10x+169
f′(x)=x²−6x+25x−3+x²+10x+169x+5
f′′(x)=x²−6x+2516+x²+10x+169144
I don't quite get it bro, do I need find out if f′′(1)>0?
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I think you are almost done. Let A(3,−4),B(−5,12),P(x,0), then AP+BP≥AB by triangle inequality, and the equality holds when P lies on segment AB.
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−4 instead of +4?
Why is it that point A has an ordinate ofLog in to reply
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−∞? @Vincent Lopez
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@Percy Jackson it's 1 but I don't know how to solve it.
P.S edited already, should be help.