Question

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4(2x-5)^2 - (3x+1)^2 =0 aflati-l pe x

Answer 1

$4 {(2x - 5)}^{2} - {(3x + 1)}^{2} = 0$

$4(4 {x}^{2} - 20x + 25) - (9 {x}^{2} + 6x + 1) = 0$

$16 {x}^{2} - 80x + 100 - 9 {x}^{2} - 6x - 1 = 0$

$7 {x}^{2} - 86x + 99 = 0$

$a = 7$

$b = - 86$

$c = 99$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 86)}^{2} - 4 \times 7 \times 99$

$\Delta = 7396 - 2772$

$\Delta = 4624$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$

$x_{1,2}=\frac{-( - 86)\pm\sqrt{4624}}{2 \times 7}$

$x_{1,2}=\frac{86\pm68}{14}$

$x_{1}=\frac{86 + 68}{14}$

$x_{1}=\frac{86 + 68}{14}$

$x_{1}=\frac{154}{14}$

$x_{1}=11$

$x_{2}=\frac{86 - 68}{14}$

$x_{2}=\frac{18}{14}$

$x_{2}=\frac{9}{7}$