Which shows the correct substitution of the values a, b, and c from the equation 1 = –2x + 3x2 + 1 into the quadratic formula? Quadratic formula: x = StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (3)(0) EndRoot Over 2(3) EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (3)(2) EndRoot Over 2(3) EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (3)(1) EndRoot Over 2(3) EndFraction x = StartFraction negative 3 plus or minus StartRoot 3 squared minus 4 (negative 2)(0) EndRoot Over 2(negative 2) EndFraction
Accepted Solution
A:
First of all, you have to manipulate the equation into the standard[tex]ax^2+bx+c=0[/tex]form. You can simplify the 1's on both sides and you have[tex]3x^2-2x=0[/tex]This means that your coefficients are[tex]a=3,\quad b=-2,\quad c=0[/tex]And since the solving formula is[tex]x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Plugging your values yields[tex]x_{1,2}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot 3\cdot 0}}{2\cdot 3}[/tex]