Tìm x : √1-4x+4x ² = 5

$\sqrt{1-4x+4x^2}=5$

$⇒1-4x+4x^2=25$

$⇒1-4x+4x^2-25=25-25$

$⇒4x^2-4x-24=0$

$⇒x_{1,\:2}=\dfrac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:4\left(-24\right)}}{2\cdot \:4}$

$⇒x_{1,\:2}=\dfrac{-\left(-4\right)\pm \:20}{2\cdot \:4}$

$⇒x=\dfrac{-\left(-4\right)+20}{2\cdot \:4}hoặc\:x=\dfrac{-\left(-4\right)-20}{2\cdot \:4}$

TH1:

$x=\dfrac{-\left(-4\right)+20}{2\cdot \:4}$

$x=\dfrac{4+20}{2\cdot \:4}$ $x=\dfrac{24}{2\cdot \:4}$

$x=\dfrac{2\cdot \:12}{2\cdot \:4}$ $x=\dfrac{12}{4}$

$x=3$

TH2:

$x=\dfrac{-\left(-4\right)-20}{2\cdot \:4}$

$x=\dfrac{4-20}{2\cdot \:4}$ $x=\dfrac{-16}{2\cdot \:4}$

$x=\dfrac{-2\cdot \:8}{2\cdot \:4}$ $x=\dfrac{-8}{4}$

$x=-2$

Vậy $x=3$ hoặc $x=-2$