Solve (k*-k+3/2)^2

$$( k \times - k + \frac { 3 } { 2 } ) ^ { 2 }$$

Evaluate

$$\frac{\left(3-2k^{2}\right)^{2}}{4}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k\left(-k\right)+\frac{3}{2}\right)^{2}.

$$k^{2}\left(-k\right)^{2}+3k\left(-k\right)+\frac{9}{4}$$

Calculate -k to the power of 2 and get k^{2}.

$$k^{2}k^{2}+3k\left(-k\right)+\frac{9}{4}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$k^{4}+3k\left(-k\right)+\frac{9}{4}$$

Multiply k and k to get k^{2}.

$$k^{4}+3k^{2}\left(-1\right)+\frac{9}{4}$$

Multiply 3 and -1 to get -3.

$$k^{4}-3k^{2}+\frac{9}{4}$$

Expand

$$k^{4}-3k^{2}+\frac{9}{4}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k\left(-k\right)+\frac{3}{2}\right)^{2}.

$$k^{2}\left(-k\right)^{2}+3k\left(-k\right)+\frac{9}{4}$$

Calculate -k to the power of 2 and get k^{2}.

$$k^{2}k^{2}+3k\left(-k\right)+\frac{9}{4}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$k^{4}+3k\left(-k\right)+\frac{9}{4}$$

Multiply k and k to get k^{2}.

$$k^{4}+3k^{2}\left(-1\right)+\frac{9}{4}$$

Multiply 3 and -1 to get -3.

$$k^{4}-3k^{2}+\frac{9}{4}$$