Answered

IDNLearn.com is designed to help you find reliable answers to any question you have. Get prompt and accurate answers to your questions from our community of knowledgeable experts.

Find the roots of the equation by completing the square: 3x^2-6x-2=0. Prove your answer by solving by the quadratic formula.

Sagot :

Luanaidn
Calculating delta: 
Δ=b²-4ac
a=3
b=-6
c=-2
Δ=36-4*3*(-2)=36+24=60
√Δ=√60
Delta is positive so there are two roots:
x1=[tex] \frac{ -b+ \sqrt[2]{delta} }{2a} [/tex]
x1=[tex] \frac{6+ \sqrt{delta} }{2*3} [/tex]=[tex] \frac{3+ \sqrt{15} }{3} [/tex]
x2=[tex] \frac{3- \sqrt{15} }{3} [/tex]
[tex]3x^2-6x-2=0\\\\(\sqrt3\ x)^2-2\cdot\sqrt3\ x\cdot\sqrt3+(\sqrt3)^2-(\sqrt3)^2-2=0\\\\(\sqrt3\ x-\sqrt3)^2-3-2=0\\\\(\sqrt3\ x-\sqrt3)^2=5\iff\sqrt3\ x-\sqrt3=-\sqrt5\ \vee\ \sqrt3\ x-\sqrt3=\sqrt5\\\\\sqrt3\ x=\sqrt3-\sqrt5\ \vee\ \sqrt3\ x=\sqrt3+\sqrt5\ \ \ \ \ |multiply\ both\ sides\ by\ \sqrt3\\\\3x=3-\sqrt{15}\ \vee\ 3x=3+\sqrt{15}\ \ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\x=\frac{3-\sqrt{15}}{3}\ \vee\ x=\frac{3+\sqrt{15}}{3}[/tex]



[tex]Prove:\\\\3x^2-6x-2=0\\a=3;\ b=-6;\ c=-2\\\Delta=b^2-4ac\to\Delta=(-6)^2-4\cdot3\cdot(-2)=36+24=60\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{60}=\sqrt{4\cdot15}=\sqrt4\cdot\sqrt{15}=2\sqrt{15}\\\\x_1=\frac{6-2\sqrt{15}}{2\cdot3}=\frac{3-\sqrt{15}}{3}\ \vee\ x_2=\frac{6+2\sqrt{15}}{2\cdot3}=\frac{3+\sqrt{15}}{3}[/tex]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com for your queries. We’re committed to providing accurate answers, so visit us again soon.