Question

I need help with this problem x^2=7x+17 step by step using the “completing the square” method

Answer 1

Answer:

x = $\frac{7-3\sqrt{13}}{2}$ x= $\frac{7-3\sqrt{13}}{2}$

Step-by-step explanation:

Step 1: Subtract 7x on both sides

$x^{2} = 7x + 17$    

Step 2: Subtract 17 on both sides

$x^{2} - 7x = 17\\x^{2} -7x - 17 = 0$

Solve with the quadratic formula:

$\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-7,\:c=-17:\quad x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:1\left(-17\right)}}{2\cdot \:1}$

$\frac{-\left(-7\right)+\sqrt{\left(-7\right)^2-4\cdot \:1\cdot \left(-17\right)}}{2\cdot \:1}$

= $\frac{7+\sqrt{117}}{2\cdot \:1}$

= $\frac{7+\sqrt{117}}{2}$

= $\frac{7+3\sqrt{13}}{2}$

$\frac{-\left(-7\right)-\sqrt{\left(-7\right)^2-4\cdot \:1\cdot \left(-17\right)}}{2\cdot \:1}$

= $\frac{7-\sqrt{117}}{2\cdot \:1}$

= $\frac{7-\sqrt{117}}{2}$

= $\frac{7-3\sqrt{13}}{2}$

x = $\frac{7-3\sqrt{13}}{2}$ x= $\frac{7-3\sqrt{13}}{2}$

Answer 2

$x^2=7x+17\\x^2-7x-17=0\\D=(-7)^2-4*(-17)=117\\\\x_1=\frac{7+\sqrt{117} }{2} =\frac{7+3\sqrt{13} }{2} \\x_2=\frac{7-\sqrt{117} }{2} =\frac{7-3\sqrt{13} }{2}$

$Answer: \\x_1=\frac{7+3\sqrt{13} }{2} \\\\x_2=\frac{7-3\sqrt{13} }{2}$

P.S. Hello from Russia