X 3 2x 2 X

$\exponential{(10)}{2} - 2 x - three $

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a+b=-two ab=1\left(-3\right)=-iii

Cistron the expression by grouping. Showtime, the expression needs to exist rewritten equally x^{2}+ax+bx-three. To discover a and b, gear up a system to be solved.

a=-3 b=1

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

\left(x^{two}-3x\right)+\left(x-3\right)

Rewrite x^{2}-2x-3 every bit \left(x^{ii}-3x\right)+\left(10-three\right).

x\left(x-3\right)+10-iii

Gene out x in x^{ii}-3x.

\left(x-3\correct)\left(x+one\correct)

Cistron out common term ten-three by using distributive property.

x^{two}-2x-3=0

Quadratic polynomial tin can exist factored using the transformation ax^{ii}+bx+c=a\left(x-x_{1}\correct)\left(x-x_{2}\correct), where x_{1} and x_{two} are the solutions of the quadratic equation ax^{2}+bx+c=0.

x=\frac{-\left(-ii\right)±\sqrt{\left(-ii\right)^{two}-iv\left(-3\right)}}{2}

All equations of the form ax^{ii}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, 1 when ± is addition and one when it is subtraction.

x=\frac{-\left(-2\correct)±\sqrt{four-four\left(-iii\correct)}}{2}

Square -2.

x=\frac{-\left(-2\right)±\sqrt{four+12}}{2}

Multiply -four times -3.

x=\frac{-\left(-2\correct)±\sqrt{16}}{two}

Add iv to 12.

x=\frac{-\left(-2\correct)±4}{2}

Have the square root of 16.

x=\frac{2±four}{2}

The opposite of -2 is 2.

10=\frac{6}{2}

At present solve the equation x=\frac{2±iv}{2} when ± is plus. Add together 2 to four.

x=\frac{-ii}{2}

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

x^{2}-2x-3=\left(x-three\right)\left(x-\left(-1\correct)\right)

Factor the original expression using ax^{two}+bx+c=a\left(10-x_{1}\right)\left(ten-x_{2}\right). Substitute iii for x_{1} and -1 for x_{two}.

10^{two}-2x-3=\left(x-3\right)\left(x+i\right)

Simplify all the expressions of the course p-\left(-q\right) to p+q.

x ^ 2 -2x -three = 0

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

r + south = 2 rs = -3

Let r and due south be the factors for the quadratic equation such that 10^2+Bx+C=(x−r)(10−southward) where sum of factors (r+s)=−B and the product of factors rs = C

r = i - u southward = 1 + u

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

(1 - u) (1 + u) = -3

To solve for unknown quantity u, substitute these in the product equation rs = -3

ane - u^two = -three

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

-u^2 = -iii-1 = -4

Simplify the expression by subtracting 1 on both sides

u^2 = four u = \pm\sqrt{4} = \pm two

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

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

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