Question

Solve each equation by completing the square. If necessary, round to the nearest hundredth.

Answer 1

$1.)\\ \\ b^2+10b=75\\ \\ b^2+10b-75=0\\ \\\Delta = b^{2}-4ac = 10^{2}-4*1*(75)=100+300=400 \\ \\\sqrt{\Delta }=\sqrt{400}=20$

$b_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-10-20}{2}=\frac{-30}{2}=-15\\ \\b_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{-10+20}{2}=\frac{10}{2}=5$



$2.)\\ \\ n^2-20n=-75\\ \\ n^2-20n+75=0\\ \\\Delta = b^{2}-4ac = (-20)^{2}-4*1*75=400-300=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$n_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{20-10}{2}=\frac{10}{2}=5\\ \\n_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{20+10}{2}=\frac{30}{2}=15$



$3.)\\ \\t^2+8t-9=0\\ \\\Delta = b^{2}-4ac = 8^{2}-4*1* (-9)=64+36=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$t_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-8-10}{2}=\frac{-18}{2}=-9\\ \\t_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{-8+10}{2}=\frac{2}{2}=1$


$4.)\\ \\m^2-2m-8=0\\ \\\Delta = b^{2}-4ac = (-2)^{2}-4*1* (-8)=4+32=36\\ \\\sqrt{\Delta }=\sqrt{36}=6$

$m_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{2-6}{2}=\frac{-4}{2}=-2\\ \\m_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{2+6}{2}=\frac{8}{2}=4$



$5.)\\ \\ v^2+4v-2=0\\ \\\Delta = b^{2}-4ac = 4^{2}-4*1* (-2)=16+8=24\\ \\\sqrt{\Delta }=\sqrt{24}= \sqrt{4*6}=2\sqrt{6}$

$v_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-4-2\sqrt{6}}{2}=\frac{2(-2-\sqrt{6}}{2}= -2-\sqrt{6}\approx -2-2,45\approx -4,45\\ \\v_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{-4+2\sqrt{6}}{2}=\frac{2(\sqrt{6}-2}{2}= \sqrt{6}-2 \approx 2,45-2\approx 0,45$