Question

Need help asap please

Answer 1

Answer:

B. x= 4

Step-by-step explanation:

We know that,

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

$UV^{2}=UX\times UY$

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,

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

i.e. $x=\frac{-12\pm \sqrt{144+256}}{2}$

i.e. $x=\frac{-12\pm \sqrt{400}}{2}$

i.e. $x=\frac{-12\pm 20}{2}$

i.e. $x=\frac{-12-20}{2}$ and i.e. $x=\frac{-12+20}{2}$

i.e. $x=\frac{-32}{2}$ and i.e. $x=\frac{8}{2}$

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

Since, length cannot be negative.

So, we have, x= 4.