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For example $5=x^2-4x+9$ , set this equation to zero by subtracting $5$ from each side. Why do we have to set it equal to zero to work it out?

Bahar
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    You don't have to. The main goal is to put it in a form $f(x)=a$ in which we know how to compute the inverse image $f^{-1}$ and apply it on both sides $x=f^{-1}(f(x))=f^{-1}(a)$. For this particular equation it turns out that that subtracting $5$ on both sides is part of computing $f^{-1}$ and incidentally that makes one side zero. But for $6=x^2-4x+9$ it is not. – Hellen Aug 16 '17 at 22:42
  • For $6=x^2-4x+9$ you can subtract $5$ and get $1=x^2-4x+4=(x-2)^2$. Then compute square root and get $\pm1=x-2$. Finally, add $2$ and get $2\pm1=x$. – Hellen Aug 16 '17 at 22:44
  • So you don't have to set it equal to 0 to solve the equation,because x is a variable? – Bahar Aug 16 '17 at 22:48
  • Edited the comment now sorry. – Bahar Aug 16 '17 at 22:51
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    The reason is not $x$ being a variable, the reason is that quadratic equations can be solved in this way. – Hellen Aug 16 '17 at 22:53

2 Answers2

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So that we could be able to factorize it and use the $AB = 0$ Principle.

The $AB = 0$ is like this, if you have two numbers with products equal to zero, then either both numbers are zero, or one of then must be zero.

Consider your equation, $x^2 - 4x + 9 = 5$ now if we can set the the $LHS$ equal to zero, then we can factorize and use the $AB = 0$ principle.

As $x^2 - 4x + 4 = 0$ then $x(x - 2) -2(x - 2) = 0$ which yields $( x - 2)(x - 2) = 0$

For equality to hold, $(x - 2)$ must be equal to $0$, which is a clever way to find the roots of a quadratic equation.

Icosahedron
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  • So you factorise the LHS not the whole equation, thank you about that. – Bahar Aug 16 '17 at 23:04
  • The equation has two sides, one side is $0$ the other contains terms and variables, we factorize the side containing terms and variables, since it is equal to zero, then we use the $AB =0$ principle. – Icosahedron Aug 17 '17 at 05:20
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You don't always have to set anything to zero. Consider the equation $x^2-4x = 12,$ which can be solved by the method of "completing the square" as follows: \begin{align} x^2-4x &= 12 \\ (x^2-4x+4) - 4 &= 12 \\ (x - 2)^2 - 4 &= 12 \\ (x - 2)^2 - 4 + 4 &= 12 + 4 \\ (x - 2)^2 &= 16 \\ x - 2 &= \pm 4 \\ \end{align} Hence either $x = 6$ or $x = -2.$

On the other hand, I can think immediately of two methods in which the first step is to gather everything on one side of the equation, leaving zero on the other side. One of these is the factorization method (already explained in another answer); the other is the usual quadratic formula, $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$ which assumes that you have put the equation into the form $ ax^2 + bx + c = 0.$ (How would you use this formula to solve $5=x^2-4x+9$ without subtracting $5$ from each side?)

David K
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