Quadratic equation - Viets

In the equation 3x²+bx+c=0, one root is x1 = -3/2. Determine the number c so that the number 4 is the root of the equation.

Hint - use Viet's formulas.

Final Answer:

c =  -18

Step-by-step explanation:

$$\begin{gathered} 3x^2-2x+c=0 \\ {x}_1=-3/2=-\frac{3}{2}=-1\frac{1}{2}=-1.5 \\ {x}_2=4 \\ a=3 \\ a(x-x_1)(x-x_2)=a\cdot x^2+b\cdot x+c \\ 3\cdot (x-x_1)\cdot (x-x_2)=3x^2+b_x+c \\ 3\cdot (x+3/2)\cdot (x-4)=3x^2+b_x+c \\ {x}_1.x_2=c/a \\ {x}_1+x_2=-b/a \\ b=-(x_1+x_2)\cdot a=-((-1.5)+4)\cdot 3=-\frac{15}{2}=-7\frac{1}{2}=-7.5 \\ c=a\cdot (x_1\cdot x_2)=3\cdot ((-1.5)\cdot 4)=-18 \\ 3\cdot x_1\cdot x_2=c \\ c=3\cdot x_1\cdot x_2=3\cdot (-1.5)\cdot 4=-18 \\ \text{ Verifying Solution: } \\ 3z^2-7.5z+c=0 \\ 3z^2-7.5z+(-18)=0 \\ 3z^2-7.5z-18=0 \\ a=3;b=-7.5;c=-18 \\ D=b^2-4ac=7.5^2-4\cdot 3\cdot (-18)=272.25 \\ D>0 \\ {z}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{7.5\pm \sqrt{272.25}}{6} \\ {z}_{1,2}=1.25\pm 2.75 \\ {z}_1=4 \\ {z}_2=-1.5 \end{gathered}$$

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