Solve {r}{x^2+(2k-1)x+4=0}{x=2}

$$\left. \begin{array} { r } { x ^ { 2 } + ( 2 k - 1 ) x + 4 = 0 } \\ { x = 2 } \end{array} \right.$$

Solve for x, k

$$x=2$$

$$k = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$2^{2}+\left(2k-1\right)\times 2+4=0$$

Calculate 2 to the power of 2 and get 4.

$$4+\left(2k-1\right)\times 2+4=0$$

Use the distributive property to multiply 2k-1 by 2.

$$4+4k-2+4=0$$

Subtract 2 from 4 to get 2.

$$2+4k+4=0$$

Add 2 and 4 to get 6.

$$6+4k=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$4k=-6$$

Divide both sides by 4.

$$k=\frac{-6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$k=-\frac{3}{2}$$

The system is now solved.

$$x=2 k=-\frac{3}{2}$$