Question

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.​

Answer 1

In both cases,

$AB^2=AD^2$

(as a consequence of the interesecting secant-tangent theorem)

So we have

10.

$(4x+7)^2=(6x-3)^2$

$16x^2+56x+49=36x^2-36x+9$

$20x^2-92x-40=0$

$5x^2-23x-10=0$

$(5x+2)(x-5)=0\implies\boxed{x=5}$

(omit the negative solution because that would make at least one of AB or AD have negative length)

11.

$(4x^2-18x-10)^2=(x^2+x+4)^2$

$16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16$

$15x^4-146x^3+235x^2+352x+84=0$

$(x-7)(3x+2)(5x^2-17x-6)=0\implies\boxed{x=-\dfrac23\text{ or }x=7}$

(again, omit the solutions that would give a negative length for either AB or AD)