Solve (1/2p-3/4q)(1/2p+3/4q)quad=

$$( \frac { 1 } { 2 } p - \frac { 3 } { 4 } q ) ( \frac { 1 } { 2 } p + \frac { 3 } { 4 } q ) \quad =$$

Evaluate

$$\frac{p^{2}}{4}-\frac{9q^{2}}{16}$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{1}{2}p\right)^{2}-\left(\frac{3}{4}q\right)^{2}$$

Expand \left(\frac{1}{2}p\right)^{2}.

$$\left(\frac{1}{2}\right)^{2}p^{2}-\left(\frac{3}{4}q\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}p^{2}-\left(\frac{3}{4}q\right)^{2}$$

Expand \left(\frac{3}{4}q\right)^{2}.

$$\frac{1}{4}p^{2}-\left(\frac{3}{4}\right)^{2}q^{2}$$

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{1}{4}p^{2}-\frac{9}{16}q^{2}$$

Expand

$$\frac{p^{2}}{4}-\frac{9q^{2}}{16}$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{1}{2}p\right)^{2}-\left(\frac{3}{4}q\right)^{2}$$

Expand \left(\frac{1}{2}p\right)^{2}.

$$\left(\frac{1}{2}\right)^{2}p^{2}-\left(\frac{3}{4}q\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}p^{2}-\left(\frac{3}{4}q\right)^{2}$$

Expand \left(\frac{3}{4}q\right)^{2}.

$$\frac{1}{4}p^{2}-\left(\frac{3}{4}\right)^{2}q^{2}$$

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{1}{4}p^{2}-\frac{9}{16}q^{2}$$