Question

Which equation has the solutions mc013-1.jpg? 3x2 – 5x + 7 = 0 3x2 – 5x – 1 = 0 3x2 – 10x + 6 = 0 3x2 – 10x – 1 = 0

Answer 1

d

Explanation:

Answer 2

Given\:solutions:
`(5+2√(7))/(3)\:and\:(5-2√(7))/(3)`.

Therefore, factors of the equation would be
`(x-(5+2√(7))/(3))\:and\:(x-(5-2√(7))/(3))`

Let us multiply those two factors to get the equation, we get

`\left(x-(5+2√(7))/(3)\right)\left(x-(5-2√(7))/(3)\right)`

`\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd`

`a=x,\:b=-(5+2√(7))/(3),\:c=x,\:d=-(5-2√(7))/(3)`

`=xx+x\left(-(5-2√(7))/(3)\right)+\left(-(5+2√(7))/(3)\right)x+\left(-(5+2√(7))/(3)\right)\left(-(5-2√(7))/(3)\right)`

`=xx-(5-2√(7))/(3)x-(5+2√(7))/(3)x+(5+2√(7))/(3)\cdot (5-2√(7))/(3)`

`=x^2-(5x-2√(7)x)/(3)-(5x+2√(7)x)/(3)-(1)/(3)`

`\mathrm{Combine\:the\:fractions\:}-(5x-2√(7)x)/(3)-(5x+2√(7)x)/(3)-(1)/(3):\quad (-\left(5x-2√(7)x\right)-\left(5x+2√(7)x\right)-1)/(3)`

`=x^2+(-\left(5x-2√(7)x\right)-\left(5x+2√(7)x\right)-1)/(3)`

`\mathrm{Expand}\:-\left(5x-2√(7)x\right)-\left(5x+2√(7)x\right)-1:\quad -10x-1`

`=x^2+(-10x-1)/(3)`

Setting it equal to 0.

`x^2+(-10x-1)/(3)=0`

Multiplying whole equation by 3, we get

`3x^2-10x-1 = 0`

Therefore, correct option is D
`3x^2-10x-1 = 0`