Solve 4p^2-25p+21=-4

$$4 p ^ { 2 } - 25 p + 21 = - 4$$

Solve for p

$$p = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$p=5$$

Steps Using Factoring By Grouping

Add 4 to both sides.

$$4p^{2}-25p+21+4=0$$

Add 21 and 4 to get 25.

$$4p^{2}-25p+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4p^{2}+ap+bp+25. To find a and b, set up a system to be solved.

$$a+b=-25 ab=4\times 25=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -25.

$$a=-20 b=-5$$

Rewrite 4p^{2}-25p+25 as \left(4p^{2}-20p\right)+\left(-5p+25\right).

$$\left(4p^{2}-20p\right)+\left(-5p+25\right)$$

Factor out 4p in the first and -5 in the second group.

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

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

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

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

$$p=5 p=\frac{5}{4}$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4p^{2}-25p+21=-4$$

Add 4 to both sides of the equation.

$$4p^{2}-25p+21-\left(-4\right)=-4-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$4p^{2}-25p+21-\left(-4\right)=0$$

Subtract -4 from 21.

$$4p^{2}-25p+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -25 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Square -25.

$$p=\frac{-\left(-25\right)±\sqrt{625-4\times 4\times 25}}{2\times 4}$$

Multiply -4 times 4.

$$p=\frac{-\left(-25\right)±\sqrt{625-16\times 25}}{2\times 4}$$

Multiply -16 times 25.

$$p=\frac{-\left(-25\right)±\sqrt{625-400}}{2\times 4}$$

Add 625 to -400.

$$p=\frac{-\left(-25\right)±\sqrt{225}}{2\times 4}$$

Take the square root of 225.

$$p=\frac{-\left(-25\right)±15}{2\times 4}$$

The opposite of -25 is 25.

$$p=\frac{25±15}{2\times 4}$$

Multiply 2 times 4.

$$p=\frac{25±15}{8}$$

Now solve the equation p=\frac{25±15}{8} when ± is plus. Add 25 to 15.

$$p=\frac{40}{8}$$

Divide 40 by 8.

$$p=5$$

Now solve the equation p=\frac{25±15}{8} when ± is minus. Subtract 15 from 25.

$$p=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$p=\frac{5}{4}$$

The equation is now solved.

$$p=5 p=\frac{5}{4}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4p^{2}-25p+21=-4$$

Subtract 21 from both sides of the equation.

$$4p^{2}-25p+21-21=-4-21$$

Subtracting 21 from itself leaves 0.

$$4p^{2}-25p=-4-21$$

Subtract 21 from -4.

$$4p^{2}-25p=-25$$

Divide both sides by 4.

$$\frac{4p^{2}-25p}{4}=-\frac{25}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$p^{2}-\frac{25}{4}p=-\frac{25}{4}$$

Divide -\frac{25}{4}, the coefficient of the x term, by 2 to get -\frac{25}{8}. Then add the square of -\frac{25}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{25}{4}p+\left(-\frac{25}{8}\right)^{2}=-\frac{25}{4}+\left(-\frac{25}{8}\right)^{2}$$

Square -\frac{25}{8} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{25}{4}p+\frac{625}{64}=-\frac{25}{4}+\frac{625}{64}$$

Add -\frac{25}{4} to \frac{625}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{25}{4}p+\frac{625}{64}=\frac{225}{64}$$

Factor p^{2}-\frac{25}{4}p+\frac{625}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-\frac{25}{8}\right)^{2}=\frac{225}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{25}{8}\right)^{2}}=\sqrt{\frac{225}{64}}$$

Simplify.

$$p-\frac{25}{8}=\frac{15}{8} p-\frac{25}{8}=-\frac{15}{8}$$

Add \frac{25}{8} to both sides of the equation.

$$p=5 p=\frac{5}{4}$$