Respuesta :

LammettHash

In both cases,

[tex]AB^2=AD^2[/tex]

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

So we have

10.

[tex](4x+7)^2=(6x-3)^2[/tex]

[tex]16x^2+56x+49=36x^2-36x+9[/tex]

[tex]20x^2-92x-40=0[/tex]

[tex]5x^2-23x-10=0[/tex]

[tex](5x+2)(x-5)=0\implies\boxed{x=5}[/tex]

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

11.

[tex](4x^2-18x-10)^2=(x^2+x+4)^2[/tex]

[tex]16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16[/tex]

[tex]15x^4-146x^3+235x^2+352x+84=0[/tex]

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

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

The value of x for first figure is x = 5 and for second x = 7 and -2/3.

What is the property of tangent?

The property of tangent is that "if two tangents from the same exterior point are tangent to a circle, then they are congruent".

1. The value of x using the above tangent property.

BA = AD

4x + 7 = 6x -3

4x - 6x = -3 -7

-2x = -10

x = -10/-2

x = 5

2. The value of x using the above tangent property.

BA = AD

[tex]\rm 4x^2-18x-10=x^2+x+4\\\\4x^2-18x-10-x^2-x-4=0\\\\3x^2-19x-14=0\\\\x =\dfrac{-(-19)\pm\sqrt{(-19)^2-4\times 3\times -14} }{2\times 3}\\\\x =\dfrac{19\pm\sqrt{361+168} }{6}\\\\x =\dfrac{19\pm\sqrt{529} }{6}\\\\x =\dfrac{19+23 }{6}, \ x =\dfrac{19-\ 23}{6}\\\\x =\dfrac{42}{6} , \ x =\dfrac{-4}{6}\\\\x=7, \ x=\dfrac{-2}{3}[/tex]

Hence, the value of x for first figure is x = 5 and for second x = 7 and -2/3.

Learn more about tangent here;

https://brainly.com/question/15571062

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