Solve 4y^2+41y+100=0

$$4 { y }^{ 2 } +41y+100 = 0$$

Solve for y

$$y = -\frac{25}{4} = -6\frac{1}{4} = -6.25$$

$$y=-4$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4y^{2}+ay+by+100. To find a and b, set up a system to be solved.

$$a+b=41 ab=4\times 100=400$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 400.

$$1,400 2,200 4,100 5,80 8,50 10,40 16,25 20,20$$

Calculate the sum for each pair.

$$1+400=401 2+200=202 4+100=104 5+80=85 8+50=58 10+40=50 16+25=41 20+20=40$$

The solution is the pair that gives sum 41.

$$a=16 b=25$$

Rewrite 4y^{2}+41y+100 as \left(4y^{2}+16y\right)+\left(25y+100\right).

$$\left(4y^{2}+16y\right)+\left(25y+100\right)$$

Factor out 4y in the first and 25 in the second group.

$$4y\left(y+4\right)+25\left(y+4\right)$$

Factor out common term y+4 by using distributive property.

$$\left(y+4\right)\left(4y+25\right)$$

To find equation solutions, solve y+4=0 and 4y+25=0.

$$y=-4 y=-\frac{25}{4}$$

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.

$$4y^{2}+41y+100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 41 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-41±\sqrt{41^{2}-4\times 4\times 100}}{2\times 4}$$

Square 41.

$$y=\frac{-41±\sqrt{1681-4\times 4\times 100}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-41±\sqrt{1681-16\times 100}}{2\times 4}$$

Multiply -16 times 100.

$$y=\frac{-41±\sqrt{1681-1600}}{2\times 4}$$

Add 1681 to -1600.

$$y=\frac{-41±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$y=\frac{-41±9}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-41±9}{8}$$

Now solve the equation y=\frac{-41±9}{8} when ± is plus. Add -41 to 9.

$$y=-\frac{32}{8}$$

Divide -32 by 8.

$$y=-4$$

Now solve the equation y=\frac{-41±9}{8} when ± is minus. Subtract 9 from -41.

$$y=-\frac{50}{8}$$

Reduce the fraction \frac{-50}{8} to lowest terms by extracting and canceling out 2.

$$y=-\frac{25}{4}$$

The equation is now solved.

$$y=-4 y=-\frac{25}{4}$$

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.

$$4y^{2}+41y+100=0$$

Subtract 100 from both sides of the equation.

$$4y^{2}+41y+100-100=-100$$

Subtracting 100 from itself leaves 0.

$$4y^{2}+41y=-100$$

Divide both sides by 4.

$$\frac{4y^{2}+41y}{4}=-\frac{100}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\frac{41}{4}y=-\frac{100}{4}$$

Divide -100 by 4.

$$y^{2}+\frac{41}{4}y=-25$$

Divide \frac{41}{4}, the coefficient of the x term, by 2 to get \frac{41}{8}. Then add the square of \frac{41}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{41}{4}y+\left(\frac{41}{8}\right)^{2}=-25+\left(\frac{41}{8}\right)^{2}$$

Square \frac{41}{8} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{41}{4}y+\frac{1681}{64}=-25+\frac{1681}{64}$$

Add -25 to \frac{1681}{64}.

$$y^{2}+\frac{41}{4}y+\frac{1681}{64}=\frac{81}{64}$$

Factor y^{2}+\frac{41}{4}y+\frac{1681}{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}.

$$\left(y+\frac{41}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{41}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$y+\frac{41}{8}=\frac{9}{8} y+\frac{41}{8}=-\frac{9}{8}$$

Subtract \frac{41}{8} from both sides of the equation.

$$y=-4 y=-\frac{25}{4}$$