Solve (a+1/2b)(a-1/2b)-(3a-2b)

$$( a + \frac { 1 } { 2 } b ) ( a - \frac { 1 } { 2 } b ) - ( 3 a - 2 b )$$

Evaluate

$$-\frac{b^{2}}{4}+a^{2}+2b-3a$$

Solution Steps

Consider \left(a+\frac{1}{2}b\right)\left(a-\frac{1}{2}b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$a^{2}-\left(\frac{1}{2}b\right)^{2}-\left(3a-2b\right)$$

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

$$a^{2}-\left(\frac{1}{2}\right)^{2}b^{2}-\left(3a-2b\right)$$

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

$$a^{2}-\frac{1}{4}b^{2}-\left(3a-2b\right)$$

To find the opposite of 3a-2b, find the opposite of each term.

$$a^{2}-\frac{1}{4}b^{2}-3a-\left(-2b\right)$$

The opposite of -2b is 2b.

$$a^{2}-\frac{1}{4}b^{2}-3a+2b$$

Expand

$$-\frac{b^{2}}{4}+a^{2}+2b-3a$$

Solution Steps

Consider \left(a+\frac{1}{2}b\right)\left(a-\frac{1}{2}b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$a^{2}-\left(\frac{1}{2}b\right)^{2}-\left(3a-2b\right)$$

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

$$a^{2}-\left(\frac{1}{2}\right)^{2}b^{2}-\left(3a-2b\right)$$

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

$$a^{2}-\frac{1}{4}b^{2}-\left(3a-2b\right)$$

To find the opposite of 3a-2b, find the opposite of each term.

$$a^{2}-\frac{1}{4}b^{2}-3a-\left(-2b\right)$$

The opposite of -2b is 2b.

$$a^{2}-\frac{1}{4}b^{2}-3a+2b$$