Respuesta :

Answer:

So, The roots are [tex]x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}[/tex]

Step-by-step explanation:

[tex]4x^2 + 5x + 2 = 2x^2 + 7x - 1[/tex]

We need to solve the equation to find the roots using quadratic formula.

The quadratic formula is:

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

Rearranging the above equation:

[tex]4x^2 -2x^2+ 5x-7x + 2+1 =0[/tex]

[tex]2x^2 -2x + 3 =0[/tex]

Where a =2 , b=-2 and c =3 Putting values in quadratic equation and solving:

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(3)}}{2(2)}\\x=\frac{2\pm\sqrt{4-24}}{4}\\x=\frac{2\pm\sqrt{-20}}{4}\\\sqrt{-20} \,\,can\,\,be\,\, written\,\, as\,\, 2\sqrt{-5}\\ x=\frac{2\pm2\sqrt{-5}}{4}\\x=\frac{2+2\sqrt{-5}}{4} \,\, and \,\, x=\frac{2-2\sqrt{-5}}{4}\\x=\frac{2(1+\sqrt{-5})}{4} \,\, and \,\, x=\frac{2(1-\sqrt{-5})}{4}\\ x=\frac{1+\sqrt{-5}}{2} \,\, and \,\, x=\frac{1-\sqrt{-5}}{2}\\As \,\,we\,\, know\,\, \sqrt{-1} = i \\[/tex]

[tex]x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}[/tex]

So, The roots are [tex]x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}[/tex]

Answer:

these numbers go in the spaces. short and simple.

1 5

2 2

mark brainliest please!

Step-by-step explanation:

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