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AC

log

ln

(

)

7

8

9

τ

π

4

5

6

≤

≥

%

θ

1

2

3

<

>

x

i

0

.

y

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Type a math problem

AC

log

ln

(

)

7

8

9

τ

π

4

5

6

≤

≥

%

θ

1

2

3

<

>

x

i

0

.

y

Solve for x

x=-1<br/>x=5

$x=−1$

$x=5$

$x=5$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Direct Factoring Method

Steps Using Factoring

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

$x_{2}−4x−5=0$

To solve the equation, factor x^{2}-4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

To solve the equation, factor $x_{2}−4x−5$ using formula $x_{2}+(a+b)x+ab=(x+a)(x+b)$. To find $a$ and $b$, set up a system to be solved.

a+b=-4 ab=-5

$a+b=−4$ $ab=−5$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

a=-5 b=1

$a=−5$ $b=1$

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

Rewrite factored expression $(x+a)(x+b)$ using the obtained values.

\left(x-5\right)\left(x+1\right)

$(x−5)(x+1)$

To find equation solutions, solve x-5=0 and x+1=0.

To find equation solutions, solve $x−5=0$ and $x+1=0$.

x=5 x=-1

$x=5$ $x=−1$

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a+b=-4 ab=-5

To solve the equation, factor x^{2}-4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

a=-5 b=1

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

\left(x-5\right)\left(x+1\right)

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

x=5 x=-1

To find equation solutions, solve x-5=0 and x+1=0.

a+b=-4 ab=1\left(-5\right)=-5

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

a=-5 b=1

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

\left(x^{2}-5x\right)+\left(x-5\right)

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

x\left(x-5\right)+x-5

Factor out x in x^{2}-5x.

\left(x-5\right)\left(x+1\right)

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

x=5 x=-1

To find equation solutions, solve x-5=0 and x+1=0.

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

x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-5\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2}

Square -4.

x=\frac{-\left(-4\right)±\sqrt{16+20}}{2}

Multiply -4 times -5.

x=\frac{-\left(-4\right)±\sqrt{36}}{2}

Add 16 to 20.

x=\frac{-\left(-4\right)±6}{2}

Take the square root of 36.

x=\frac{4±6}{2}

The opposite of -4 is 4.

x=\frac{10}{2}

Now solve the equation x=\frac{4±6}{2} when ± is plus. Add 4 to 6.

x=5

Divide 10 by 2.

x=\frac{-2}{2}

Now solve the equation x=\frac{4±6}{2} when ± is minus. Subtract 6 from 4.

x=-1

Divide -2 by 2.

x=5 x=-1

The equation is now solved.

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

x^{2}-4x-5-\left(-5\right)=-\left(-5\right)

Add 5 to both sides of the equation.

x^{2}-4x=-\left(-5\right)

Subtracting -5 from itself leaves 0.

x^{2}-4x=5

Subtract -5 from 0.

x^{2}-4x+\left(-2\right)^{2}=5+\left(-2\right)^{2}

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

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

Square -2.

x^{2}-4x+4=9

Add 5 to 4.

\left(x-2\right)^{2}=9

Factor x^{2}-4x+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(x-2\right)^{2}}=\sqrt{9}

Take the square root of both sides of the equation.

x-2=3 x-2=-3

Simplify.

x=5 x=-1

Add 2 to both sides of the equation.

x ^ 2 -4x -5 = 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 = -5

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) = -5

To solve for unknown quantity u, substitute these in the product equation rs = -5

4 - u^2 = -5

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

-u^2 = -5-4 = -9

Simplify the expression by subtracting 4 on both sides

u^2 = 9 u = \pm\sqrt{9} = \pm 3

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

r =2 - 3 = -1 s = 2 + 3 = 5

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

$x_{2}−4x−5=0$

Trigonometry

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

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

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]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

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

${8x+2y=467x+3y=47 $

Differentiation

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

$dxd (x−5)(3x_{2}−2) $

Integration

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

$∫_{0}xe_{−x_{2}}dx$

Limits

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

$x→−3lim x_{2}+2x−3x_{2}−9 $

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