Solve (1-1/2a)^2+8(a-1/4)^2+(3/2a+1)(3/2a-1)+5a

$$( 1 - \frac { 1 } { 2 } a ) ^ { 2 } + 8 ( a - \frac { 1 } { 4 } ) ^ { 2 } + ( \frac { 3 } { 2 } a + 1 ) ( \frac { 3 } { 2 } a - 1 ) + 5 a$$

Evaluate

$$\frac{21a^{2}+1}{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(1-\frac{1}{2}a\right)^{2}.

$$1-a+\frac{1}{4}a^{2}+8\left(a-\frac{1}{4}\right)^{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-\frac{1}{4}\right)^{2}.

$$1-a+\frac{1}{4}a^{2}+8\left(a^{2}-\frac{1}{2}a+\frac{1}{16}\right)+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Use the distributive property to multiply 8 by a^{2}-\frac{1}{2}a+\frac{1}{16}.

$$1-a+\frac{1}{4}a^{2}+8a^{2}-4a+\frac{1}{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Combine \frac{1}{4}a^{2} and 8a^{2} to get \frac{33}{4}a^{2}.

$$1-a+\frac{33}{4}a^{2}-4a+\frac{1}{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Combine -a and -4a to get -5a.

$$1-5a+\frac{33}{4}a^{2}+\frac{1}{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Add 1 and \frac{1}{2} to get \frac{3}{2}.

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Consider \left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\left(\frac{3}{2}a\right)^{2}-1+5a$$

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

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\left(\frac{3}{2}\right)^{2}a^{2}-1+5a$$

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

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\frac{9}{4}a^{2}-1+5a$$

Combine \frac{33}{4}a^{2} and \frac{9}{4}a^{2} to get \frac{21}{2}a^{2}.

$$\frac{3}{2}-5a+\frac{21}{2}a^{2}-1+5a$$

Subtract 1 from \frac{3}{2} to get \frac{1}{2}.

$$\frac{1}{2}-5a+\frac{21}{2}a^{2}+5a$$

Combine -5a and 5a to get 0.

$$\frac{1}{2}+\frac{21}{2}a^{2}$$

Expand

$$\frac{21a^{2}+1}{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(1-\frac{1}{2}a\right)^{2}.

$$1-a+\frac{1}{4}a^{2}+8\left(a-\frac{1}{4}\right)^{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-\frac{1}{4}\right)^{2}.

$$1-a+\frac{1}{4}a^{2}+8\left(a^{2}-\frac{1}{2}a+\frac{1}{16}\right)+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Use the distributive property to multiply 8 by a^{2}-\frac{1}{2}a+\frac{1}{16}.

$$1-a+\frac{1}{4}a^{2}+8a^{2}-4a+\frac{1}{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Combine \frac{1}{4}a^{2} and 8a^{2} to get \frac{33}{4}a^{2}.

$$1-a+\frac{33}{4}a^{2}-4a+\frac{1}{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Combine -a and -4a to get -5a.

$$1-5a+\frac{33}{4}a^{2}+\frac{1}{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Add 1 and \frac{1}{2} to get \frac{3}{2}.

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right)+5a$$

Consider \left(\frac{3}{2}a+1\right)\left(\frac{3}{2}a-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\left(\frac{3}{2}a\right)^{2}-1+5a$$

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

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\left(\frac{3}{2}\right)^{2}a^{2}-1+5a$$

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

$$\frac{3}{2}-5a+\frac{33}{4}a^{2}+\frac{9}{4}a^{2}-1+5a$$

Combine \frac{33}{4}a^{2} and \frac{9}{4}a^{2} to get \frac{21}{2}a^{2}.

$$\frac{3}{2}-5a+\frac{21}{2}a^{2}-1+5a$$

Subtract 1 from \frac{3}{2} to get \frac{1}{2}.

$$\frac{1}{2}-5a+\frac{21}{2}a^{2}+5a$$

Combine -5a and 5a to get 0.

$$\frac{1}{2}+\frac{21}{2}a^{2}$$