Solve -1/2a^2+5/2a-2/4-a=4/2

$$\frac{ - \frac{ 1 }{ 2 } { a }^{ 2 } + \frac{ 5 }{ 2 } a-2 }{ 4-a } = \frac{ 4 }{ 2 }$$

Solve for a

$$a=5$$

Steps Using Factoring

Variable a cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 2\left(a-4\right), the least common multiple of 4-a,2.

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

Use the distributive property to multiply -2 by -\frac{1}{2}a^{2}+\frac{5}{2}a-2.

$$a^{2}-5a+4=\left(a-4\right)\times 4$$

Use the distributive property to multiply a-4 by 4.

$$a^{2}-5a+4=4a-16$$

Subtract 4a from both sides.

$$a^{2}-5a+4-4a=-16$$

Combine -5a and -4a to get -9a.

$$a^{2}-9a+4=-16$$

Add 16 to both sides.

$$a^{2}-9a+4+16=0$$

Add 4 and 16 to get 20.

$$a^{2}-9a+20=0$$

To solve the equation, factor a^{2}-9a+20 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-9 ab=20$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-5\right)\left(a-4\right)$$

To find equation solutions, solve a-5=0 and a-4=0.

$$a=5 a=4$$

Variable a cannot be equal to 4.

$$a=5$$

Steps Using Factoring By Grouping

Variable a cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 2\left(a-4\right), the least common multiple of 4-a,2.

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

Use the distributive property to multiply -2 by -\frac{1}{2}a^{2}+\frac{5}{2}a-2.

$$a^{2}-5a+4=\left(a-4\right)\times 4$$

Use the distributive property to multiply a-4 by 4.

$$a^{2}-5a+4=4a-16$$

Subtract 4a from both sides.

$$a^{2}-5a+4-4a=-16$$

Combine -5a and -4a to get -9a.

$$a^{2}-9a+4=-16$$

Add 16 to both sides.

$$a^{2}-9a+4+16=0$$

Add 4 and 16 to get 20.

$$a^{2}-9a+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+20. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 20=20$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite a^{2}-9a+20 as \left(a^{2}-5a\right)+\left(-4a+20\right).

$$\left(a^{2}-5a\right)+\left(-4a+20\right)$$

Factor out a in the first and -4 in the second group.

$$a\left(a-5\right)-4\left(a-5\right)$$

Factor out common term a-5 by using distributive property.

$$\left(a-5\right)\left(a-4\right)$$

To find equation solutions, solve a-5=0 and a-4=0.

$$a=5 a=4$$

Variable a cannot be equal to 4.

$$a=5$$

Steps Using the Quadratic Formula

Variable a cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 2\left(a-4\right), the least common multiple of 4-a,2.

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

Use the distributive property to multiply -2 by -\frac{1}{2}a^{2}+\frac{5}{2}a-2.

$$a^{2}-5a+4=\left(a-4\right)\times 4$$

Use the distributive property to multiply a-4 by 4.

$$a^{2}-5a+4=4a-16$$

Subtract 4a from both sides.

$$a^{2}-5a+4-4a=-16$$

Combine -5a and -4a to get -9a.

$$a^{2}-9a+4=-16$$

Add 16 to both sides.

$$a^{2}-9a+4+16=0$$

Add 4 and 16 to get 20.

$$a^{2}-9a+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 20}}{2}$$

Square -9.

$$a=\frac{-\left(-9\right)±\sqrt{81-4\times 20}}{2}$$

Multiply -4 times 20.

$$a=\frac{-\left(-9\right)±\sqrt{81-80}}{2}$$

Add 81 to -80.

$$a=\frac{-\left(-9\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$a=\frac{-\left(-9\right)±1}{2}$$

The opposite of -9 is 9.

$$a=\frac{9±1}{2}$$

Now solve the equation a=\frac{9±1}{2} when ± is plus. Add 9 to 1.

$$a=\frac{10}{2}$$

Divide 10 by 2.

$$a=5$$

Now solve the equation a=\frac{9±1}{2} when ± is minus. Subtract 1 from 9.

$$a=\frac{8}{2}$$

Divide 8 by 2.

$$a=4$$

The equation is now solved.

$$a=5 a=4$$

Variable a cannot be equal to 4.

$$a=5$$

Steps for Completing the Square

Variable a cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 2\left(a-4\right), the least common multiple of 4-a,2.

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

Use the distributive property to multiply -2 by -\frac{1}{2}a^{2}+\frac{5}{2}a-2.

$$a^{2}-5a+4=\left(a-4\right)\times 4$$

Use the distributive property to multiply a-4 by 4.

$$a^{2}-5a+4=4a-16$$

Subtract 4a from both sides.

$$a^{2}-5a+4-4a=-16$$

Combine -5a and -4a to get -9a.

$$a^{2}-9a+4=-16$$

Subtract 4 from both sides.

$$a^{2}-9a=-16-4$$

Subtract 4 from -16 to get -20.

$$a^{2}-9a=-20$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-9a+\left(-\frac{9}{2}\right)^{2}=-20+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-9a+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$a^{2}-9a+\frac{81}{4}=\frac{1}{4}$$

Factor a^{2}-9a+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$a-\frac{9}{2}=\frac{1}{2} a-\frac{9}{2}=-\frac{1}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$a=5 a=4$$

Variable a cannot be equal to 4.

$$a=5$$