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Since both lines are tangent to the circle, we have that both have the same length, then, we can write the following equaiton:
[tex]4x^2^{}+5x=5x+16[/tex]solving for x, we get:
[tex]\begin{gathered} 4x^2+5x=5x+16 \\ \Rightarrow4x^2+5x-5x=16 \\ \Rightarrow4x^2=16 \end{gathered}[/tex]dividing both sides by 4, we get the following:
[tex]\begin{gathered} \frac{1}{4}(4x^2=16) \\ \Rightarrow\frac{4}{4}x^2=\frac{16}{4} \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm\sqrt[]{4}=\pm2 \\ x=\pm2 \end{gathered}[/tex]therefore, x = 2 or x = -2.
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Rewritten by : Barada
