Question

$(tex) \huge \mathfrak \red {question}$
THE AREA OF A CIRCLE IS πx²
+ 10π x + 25π FIND THE RADIUS OF THE CIRCLE.
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Answer 1

Area of circle

$\sf \pi x^2+10 \pi x +25\pi$

• We have to find the radius of circle

Area of circle is given the quadratic form

now first find the value of x

By Quadratic formula

$\boxed{\sf \ x= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

Where ,

a= coefficient of $\sf x^2$

b= coefficient of x

c= constant term

$\sf \dag\ \ \ \pi x^2+10\pi x+ 25 \pi \\ \\ \sf\bullet a= \pi \ \ \bullet b= 10\pi \ \ \bullet \ c= 25 \pi$

Put the values in the formula

$\mapsto\sf x= \dfrac{ -10\pi \pm \sqrt{(-10\pi )^2-4\times \pi \times 25\pi}}{2\times \pi}\\ \\ \\ \mapsto\sf x= \dfrac{-10\pi \pm \sqrt{100\pi^2 - 100\pi^2}}{2\pi}\\ \\ \\ \mapsto\sf x= \dfrac{-10 \pi \pm\sqrt{ 0}}{ 2\pi }\\ \\ \\ \mapsto\sf x= \dfrac{-10\pi \pm 0}{2 \pi}\\ \\ \\ \mapsto\sf x= \dfrac{-10\pi}{2\pi}+\dfrac{0}{2\pi} \ , \ \ \dfrac{-10\pi}{2\pi}-\dfrac{0}{2\pi}\\ \\ \\ \mapsto\sf x= \cancel{\dfrac{-10\pi}{2\pi}}\ \ \cancel{\dfrac{-10\pi}{2\pi}}\\ \\ \\ \mapsto\sf x= -5 ,-5 \\ \\ \mapsto\sf x= -5$

• Now the Area of circle

$\sf \pi x^2+10\pi x+25\pi \\ \\ \mapsto\sf \pi( x^2+10x+25)\\ \\ \mapsto\sf \pi(x^2+5x+5x+25)\\ \\ \mapsto\sf \pi [x(x+5)+5(x+5)]\\ \\ \mapsto\sf \pi [(x+5)(x+5)]\\ \\ \sf\bullet\ \ Put \ value \ of \ x\\ \\ \mapsto\sf \pi[(-5+5)(-5+5)]\\ \\ \mapsto\sf \pi(0)\\ \\ \mapsto\sf Area \ of \ circle = 0$

Now find the radius of circle

$\boxed{\bigstar{\sf\ Area \ of \ circle = \pi r^2}}$

$\mapsto\sf \pi r^2= 0\\ \\ \mapsto\sf r^2= \dfrac{0}{\pi}\\ \\ \mapsto\sf r= \sqrt{0}\\ \\ \mapsto\sf r= 0 \ unit$

$\underline{\bigstar{\sf\ Radius\ of \ circle= 0}}$

Answer 2

GIVEN:

• Area of circle = πx² + 10πx + 25π

TO FIND :

• Radius of circle

SOLUTION:

Area of circle = πx² + 10πx + 25π

It is in the form of a quadratic polynomial

Finding the roots by quadratic formula

→ x = - b ±√b² - 4ac/2a

Here,

• a = π
• b = 10π
• c = 25π

Substituting the values

→ x = -10π ± √(10π)² - 4(π)(25π)/2(π)

→ x = -10π ± √100π² - 100π²/2π

→ x = - 10π/2π

→ x = - 5

Substituting the value of x in area of circle

→ πx² + 10πx + 25π

→ π(-5)² + 10(-5)π + 25π

→ 25π - 50π + 25π

→ 50π - 50π

→ 0

Area of circle = 0

Finding radius by using area of circle = πr²

πr² = 0

→ r² = 0/π

→ r = 0

.°. Hence, radius = 0 m