Question

Describe how to derive the quadratic formula from a quadratic equation in standard form

Answer 1

Answer:

The quadratic formula is derived from a quadratic equation in standard form when solving for x by completing the square. The steps involve creating a perfect square trinomial, isolating the trinomial, and taking the square root of both sides. The variable is then isolated to give the solutions to the equation.

Step-by-step explanation:

Answer 2

Answer:

$x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

Step-by-step explanation:

The standard form of a quadratic equation in x can be written as;

$ax^{2} +bx+c=0$

where a,b, and c are constants.

The first step is to subtract c on both sides of the equation;

$ax^{2} +bx=-c$

The next step is to divide both sides of the equation by the constant a;

$x^{2} +\frac{b}{a}x=-\frac{c}{a}$

Next, we complete the square on the left hand side of the equation by determining another constant c as;

$c=(\frac{b}{2a}) ^{2}=\frac{b^{2} }{4a^{2} }$

We then add this constant on both sides of the equation in order to complete the square on the L.H.S;

$x^{2} +\frac{b}{a}x+\frac{b^{2} }{4a^{2} }=\frac{b^{2} }{4a^{2} }-\frac{c}{a}$

The expression on the L.H.S of the equation is now a perfect square and can be factorized to yield;

$(x+\frac{b}{2a})^{2}=\frac{b^{2} -4ac}{4a^{2} }$

We then take square roots on both sides of the equation and simplify the expression on the R.H.S;

$(x+\frac{b}{2a} )=+/-\frac{\sqrt{b^{2}-4ac } }{2a}$

The final step is to make x the subject of the formula and a little simplification which will yield the quadratic formula;

$x=-\frac{b}{2a}+/-\frac{\sqrt{b^{2} -4ac} }{2a}$

Putting the expression on the R.H.S under a common denominator yields;

$x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

which is the quadratic formula