Question

Solve the equation 4x^3 + 4x^2-x-1 = 0 given that -1/2 is a zero of f(x) = 4x^3 + 4x^2-x-1.

Answer 1

Answer:

The solution of  equation 4x^3 + 4x^2-x-1 = 0 given that -1/2 is a zero of f(x) = 4x^3 + 4x^2-x-1 is $\frac{-1}{2} \text { and } \frac{1}{2} \text { and }-1$

Solution:

Given, cubic equation $4 x^{3}+4 x^{2}-x-1=0$

And $\frac{-1}{2}$ is a zero of $f(x)=4 x^{3}+4 x^{2}-x-1$

We have to find the other two roots of the given quadratic equation.

Let the other two roots be a, b.

Now, we know that, sum of roots of cubic equation $=\frac{-x^{2} \text { coefficient }}{x^{3} \text { coefficient }}$

$\text { Then, } \frac{-1}{2}+a+b=\frac{-4}{4} \rightarrow a+b=-1+\frac{1}{2} \rightarrow a+b=\frac{-1}{2} \rightarrow(1)$

Now, product of roots of cubic equation $=\frac{-\text { constant value }}{x^{3} \text { coefficient }}$

$\begin{array}{l}{\text { Then, } \frac{-1}{2} \times a \times b=\frac{-(-1)}{4} \rightarrow \frac{-1}{2} \times a b=\frac{1}{4}} \\\\ {\rightarrow a b=-2 \times \frac{1}{4} \rightarrow a b=\frac{-1}{2}}\end{array}$

$\text { Now we know that, }(a-b)^{2}=(a+b)^{2}-4 a b$

substitute above value in this formula

$(a-b)^{2}=\left(\frac{-1}{2}\right)^{2}-4 \times\left(\frac{-1}{2}\right)$

$(a-b)^{2}=1 / 4+2 \rightarrow(a-b)^{2}=\frac{2 \times 4+1}{4} \rightarrow(a-b)^{2}=\frac{9}{4} \rightarrow a-b=\frac{3}{2} \rightarrow(2)$

Now, solve (1) and (2)

$\begin{array}{l}{2 a=\frac{3-1}{2}} \\ {2 a=1} \\ {a=\frac{1}{2}}\end{array}$

substitute "a" value in (1)

$\frac{1}{2}+b= \frac{-1}{2} \rightarrow b= \frac{-1}{2} + \frac{-1}{2} \rightarrow b=-1$

Hence, the roots of the given cubic equation are $\frac{-1}{2} \text { and } \frac{1}{2} \text { and }-1$