Question

F(x)=x+3/2x^2-6 Find the zeros of the function algebraically

Answer 1

To find the zeros of a function f(x), we set f(x)=0 and then solve for x

Here. we have

`\begin{gathered} f(x)=x+(3)/(2)x^2-6 \\ \text{ setting }f(x)=0 \\ \Rightarrow(3)/(2)x^2+x-6=0 \\ \text{ multiplying each term by 2} \\ \Rightarrow2((3)/(2)x^2)+2(x)-2(6)=0 \\ \Rightarrow3x^2+2x-12=0 \\ \Rightarrow3x^2+2x=12 \\ \text{ Dividing each term by 3} \\ \Rightarrow x^2+(2)/(3)x=4 \\ \text{ Adding (}(1)/(2)*(2)/(3))^2\text{ =}(1)/(9)\text{to both sides;} \\ \Rightarrow x^2+(2)/(3)x+(1)/(9)=4+(1)/(9) \\ \Rightarrow(x+(1)/(3))^2=(37)/(9) \\ \Rightarrow(x+(1)/(3))=\frac{\pm\sqrt[]{37}}{3} \\ \Rightarrow x=-(1)/(3)\pm\frac{\sqrt[]{37}}{3} \\ \text{ The zeros of f(x) is given as;} \\ \Rightarrow x=-(1)/(3)+\frac{\sqrt[]{37}}{3}\text{ or }x=-(1)/(3)-\frac{\sqrt[]{37}}{3} \end{gathered}`