How can we make it easy to solve

Coradossi7xod

Coradossi7xod

Answered question

2022-03-24

How can we make it easy to solve quadratics without the Quadratic Formula?
For example,
5x27x2=0
Then, we only need to find is when
5x27x=2
x(5x7)=2
Since that when we get 22=0.
Is there any way we can find them easily without solving quadratics using formula?

Answer & Explanation

Jadyn Gentry

Jadyn Gentry

Beginner2022-03-25Added 12 answers

Unfortunately, your strategy does not help that much because it is very difficult to solve
x(5x7)=2
without simply returning to the original equation. This is because there is a 2 on the RHS. The reason we want to be able to write a quadratic in the form
(xp)(xq)=0
is so that we can exploit the zero-product property: if a×b=0, then a=0 or b=0. Once we have written an equation in the form (xp)(xq)=0, we can still instantly deduce that x=p or x=q. This is something we can't do when we have something like
x(5x7)=2,.
Instead, I would suggest dividing through by 5, which yields
x275x25=0,.
Then, add 2/5 to both sides:
x275x=25    (*) 
The LHS is almost a perfect square. Note that
(x710)2=x275x+49100,.
If we subtract 49/100 from this equation, we get
(x710)249100=x275x,.
Hence, the equation (*) can be rewritten as
(x710)249100=25,.
Add 49/100 to both sides:
(x710)2=25+49100=89100,.
Take the square root of both sides:
x710=±89100,.
Add 7/10 to both sides:
x=710±89100,,
and we are done. This method of solving quadratics is known as completing the square. Unlike factorisation, it can be used to solve any quadratic. In fact, the quadratic formula comes from completing the square on the general quadratic equation
ax2+bx+c=0,.

Marquis Ibarra

Marquis Ibarra

Beginner2022-03-26Added 9 answers

All ways of solving quadratic equations are essentially the same, but here's another tack:
You can "depress" the quadratic ax2+bx+c using the substitution x=yb2a. This substitution makes the linear term disappear. Then it's easy to solve for y and then plug that answer into the substitution and recover x.
For your example: 5x27x2=0, let x=y725=y+710. The equation becomes 5(y+710)27(y+710)2=0
which simplifies to
5y28920=0.
Solve for y (which is easy because there's no linear term):
y2=89100
y=±8910.
Which means
x=±8910+710.
So all the steps are easy and obvious as long as you remember the substitution x=yb2a.

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