Solve for b

b=2

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a+b=-4 ab=4

To solve the equation, factor b^{2}-4b+4 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.

-1,-4 -2,-2

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 4.

-1-4=-5 -2-2=-4

Calculate the sum for each pair.

a=-2 b=-2

The solution is the pair that gives sum -4.

\left(b-2\right)\left(b-2\right)

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

\left(b-2\right)^{2}

Rewrite as a binomial square.

b=2

To find equation solution, solve b-2=0.

a+b=-4 ab=1\times 4=4

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

-1,-4 -2,-2

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 4.

-1-4=-5 -2-2=-4

Calculate the sum for each pair.

a=-2 b=-2

The solution is the pair that gives sum -4.

\left(b^{2}-2b\right)+\left(-2b+4\right)

Rewrite b^{2}-4b+4 as \left(b^{2}-2b\right)+\left(-2b+4\right).

b\left(b-2\right)-2\left(b-2\right)

Factor out b in the first and -2 in the second group.

\left(b-2\right)\left(b-2\right)

Factor out common term b-2 by using distributive property.

\left(b-2\right)^{2}

Rewrite as a binomial square.

b=2

To find equation solution, solve b-2=0.

b^{2}-4b+4=0

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.

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

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

b=\frac{-\left(-4\right)±\sqrt{16-4\times 4}}{2}

Square -4.

b=\frac{-\left(-4\right)±\sqrt{16-16}}{2}

Multiply -4 times 4.

b=\frac{-\left(-4\right)±\sqrt{0}}{2}

Add 16 to -16.

b=-\frac{-4}{2}

Take the square root of 0.

b=\frac{4}{2}

The opposite of -4 is 4.

b=2

Divide 4 by 2.

b^{2}-4b+4=0

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.

\left(b-2\right)^{2}=0

Factor b^{2}-4b+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}.

\sqrt{\left(b-2\right)^{2}}=\sqrt{0}

Take the square root of both sides of the equation.

b-2=0 b-2=0

Simplify.

b=2 b=2

Add 2 to both sides of the equation.

b=2

The equation is now solved. Solutions are the same.

x ^ 2 -4x +4 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

r + s = 4 rs = 4

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = 2 - u s = 2 + u

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(2 - u) (2 + u) = 4

To solve for unknown quantity u, substitute these in the product equation rs = 4

4 - u^2 = 4

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 4-4 = 0

Simplify the expression by subtracting 4 on both sides

u^2 = 0 u = 0

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r = s = 2

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

Linear equation

y = 3x + 4

Arithmetic

699 * 533

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}