Solve 7/8(k-1)x^2+4x+1=0

$$\frac { 7 } { 8 } ( k - 1 ) x ^ { 2 } + 4 x + 1 = 0$$

Solve for k

$$k=1-\frac{32}{7x}-\frac{8}{7x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{7}{8} by k-1.

$$\left(\frac{7}{8}k-\frac{7}{8}\right)x^{2}+4x+1=0$$

Use the distributive property to multiply \frac{7}{8}k-\frac{7}{8} by x^{2}.

$$\frac{7}{8}kx^{2}-\frac{7}{8}x^{2}+4x+1=0$$

Add \frac{7}{8}x^{2} to both sides. Anything plus zero gives itself.

$$\frac{7}{8}kx^{2}+4x+1=\frac{7}{8}x^{2}$$

Subtract 4x from both sides.

$$\frac{7}{8}kx^{2}+1=\frac{7}{8}x^{2}-4x$$

Subtract 1 from both sides.

$$\frac{7}{8}kx^{2}=\frac{7}{8}x^{2}-4x-1$$

The equation is in standard form.

$$\frac{7x^{2}}{8}k=\frac{7x^{2}}{8}-4x-1$$

Divide both sides by \frac{7}{8}x^{2}.

$$\frac{8\times \frac{7x^{2}}{8}k}{7x^{2}}=\frac{8\left(\frac{7x^{2}}{8}-4x-1\right)}{7x^{2}}$$

Dividing by \frac{7}{8}x^{2} undoes the multiplication by \frac{7}{8}x^{2}.

$$k=\frac{8\left(\frac{7x^{2}}{8}-4x-1\right)}{7x^{2}}$$

Divide \frac{7x^{2}}{8}-4x-1 by \frac{7}{8}x^{2}.

$$k=1-\frac{32x+8}{7x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(\sqrt{2\left(39-7k\right)}-8\right)}{7\left(k-1\right)}\text{; }x=-\frac{2\left(\sqrt{2\left(39-7k\right)}+8\right)}{7\left(k-1\right)}\text{, }&k\neq 1\\x=-\frac{1}{4}\text{, }&k=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{2\left(\sqrt{2\left(39-7k\right)}-8\right)}{7\left(k-1\right)}\text{; }x=-\frac{2\left(\sqrt{2\left(39-7k\right)}+8\right)}{7\left(k-1\right)}\text{, }&k\neq 1\text{ and }k\leq \frac{39}{7}\\x=-\frac{1}{4}\text{, }&k=1\end{matrix}\right.$$