Question

the area of a circle is given by expression -(πx^2 + 10π x+ 25π). find the radius of circle

Answer 1

Heya@user

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Here is the answer

Here, 
area= -(πx² + 10πx + 25π)
= -πx² - 10πx - 25
so, 
a= -π
b= -10π
c= -25π
so, 
according quadratic formula,
x= (-b ± √b²-4ac)/2a
x= -(-10π) ± √{(-10π)²-4(-π)(-25π)}/2(-π)
x= 10π ± √100π² - 100π²/-2π
x= 10π ± √0/2π
x= 10π/-2π
x= -5

here given that
area= -πx² - 10πx - 25π
=π( -x² - 10x - 25)
=π{ -(-5)² - 10(-5) - 25}
=π{ -25 + (50) - 25}
=π{-25 + 50 - 25}
=π(50-50}
=π(0)
=0

Now area = 0
so,

area of circle= πr²
so, 
πr²=0
r²=0/π
r²=0
r=0

Hence, radius = 0

hope it helps to you.

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Answer 2

Answer:

$\large\underline{\bf \orange{Given :}} \\Area \: of \: circle = πx² + 10πx + 25π \\\large\underline{\bf \orange{To \: Find :}} \\Radius \: of \: the \: circle \\\large\underline{\bf\orange{Solution :}} \\\begin{gathered}\implies\underline{\boxed{ \bf Area \: of \: circle = {\pi r}^{2} }} \\ \\ \implies\sf \pi {x}^{2} + 10\pi x + 25\pi = {\pi r}^{2} \\ \\\implies\sf \cancel \pi ({x}^{2} + 10x + 25)= \cancel\pi( { r}^{2}) \\ \\ \implies\sf {r}^{2} = {x}^{2} + 10x + 25 \\ \\\implies\sf {r}^{2} = {x}^{2} + 5x + 5x + 25 \\ \\\implies\sf {r}^{2} = x( x+ 5) + 5(x + 5) \\ \\\implies\sf {r}^{2} = (x + 5)(x +5) \\ \\\implies\sf {r}^{2} =\sqrt{(x+5)^2 }\\ \\\implies\underline{\boxed{\sf r = x+5}} \end{gathered} \\ ⟹Area \: of \: circle=πr2 \\ ⟹πx2+10πx+25π=πr2 \\ ⟹π(x2+10x+25)=π(r2) \\ ⟹r2=x2+10x+25 \\ ⟹r2=x2+5x+5x+25 \\ ⟹r2=x(x+5)+5(x+5) \\ ⟹r2=(x+5)(x+5) \\ ⟹r2=(x+5)2 \\ ⟹r=x+5$