Question

Need help asap please

Answer 1

B. x= 4

Explanation:

We know that,

The result for the intersection of the secant and the tangent is given by by the figure below,

`UV^(2)=UX* UY`

According to the question, we have,

`8^(2)=x(12+x)`

i.e.
`64=12x+x^2`

i.e.
`x^2+12x-64=0`

Now, the solution of a quadratic equation
`ax^2+bx+c=0` is given by,
`x=\frac{-b\pm \sqrt{b^(2)-4ac}}{2a}`

So, we get,

`x^2+12x-64=0` implies a= 1, b= 12 and c= -64.

Thus, the solution is given by,

`x=\frac{-12\pm \sqrt{12^(2)-4* 1* (-64)}}{2* 1}`

i.e.
`x=(-12\pm √(144+256))/(2)`

i.e.
`x=(-12\pm √(400))/(2)`

i.e.
`x=(-12\pm 20)/(2)`

i.e.
`x=(-12-20)/(2)` and i.e.
`x=(-12+20)/(2)`

i.e.
`x=(-32)/(2)` and i.e.
`x=(8)/(2)`

i.e x= -16 and x= 4.

Since, length cannot be negative.

So, we have, x= 4.