Why do you sometimes finish the square (ex.[tex]x^{2} + 4x=1[/tex]) and sometimes you make one side zero(ex[tex]x^{2} -12x=-35[/tex])? How should I know the difference? Any help is appreciated! TIA

if you can split b into 'p' & 'q' (some real numbers) such that p + q = b & pq = ac ; then the quadratic equation can be easily solved by using factorisation method. Here the easiest way to guess the values of 'p' & 'q' is to find out the prime factors of 'ac' and taking any two prime numbers in random , just check whether the sum of those two primes is equal to 'b' (excluding their signs).

if you can't split b into 'p' & 'q' (some real numbers) such that p + q = b & pq = ac ; then try solving quadratic equation by completing the squares method.

For example :- x² - 12x = -35.

Lets make one side zero → x² - 12x + 35 = 0 . Now, lets break -12x into two real numbers 'p' & 'q' such that p + q = -12x & pq = 35. Prime factors of 35 are 1 , 5 & 7. If we take 5 & 7 as 'p' & 'q' then p + q = 5 + 7 = 12. but as we need -12 , the values of 'p' & 'q' are -5 & -7.

⇒ [tex]x^2 -5x - 7x + 35 = 0[/tex]

⇒ [tex]x(x - 5) - 7(x - 5) = 0[/tex]

⇒ [tex](x - 5)(x - 7) = 0[/tex]

⇒ [tex]x = 5 \: or \: 7[/tex]

Lets take another example :- x² + 4x = 1.

Lets make one side zero → x² + 4x - 1 = 0 . Now , lets split +4x into into two real numbers 'p' & 'q' such that p + q = +4x & pq = -1 . Prime factors of 4 are 1 & 2 only. But there's no such combination of its primes such that their sum would be equal to 4. So, the quadratic equation can't be solved by factorisation method. Now, lets bring -1 to Right Hand Side & try solving by completing squares method.

⇒ [tex]x^2 + 2 \times x \times 2 + 2^2 = 1 + 2^2[/tex]

⇒ [tex](x + 2)^2 = 1 + 4 = 5[/tex]

⇒ [tex]x + 2 = +\sqrt{5} \: \: or \: -\sqrt{5}[/tex]

⇒ [tex]x = (\sqrt{5} - 2) \: or \: (-\sqrt{5} -2)[/tex]

Conclusion:

First try making one side zero and try to solve the equation by factorisation method. If it got factorised , then you are done and if it didn't get factorised , then solve it by completing squares method.

janmilczek janmilczek · Answer 2 · 2025-04-01T04:00:31-04:00

Answer:

I'm assuming we're asking about the difference between:

[tex]ax^2 + bx + c = 0[/tex]

and

[tex]ax^2 + bx = -c[/tex]

Let's preface this by stating the obvious: there's no actual difference. Both equations are equivalent.

Now, it has a lot to do with our goal. Sometimes our goal is to write an equation concisely (then the "no-zero" way uses less characters), and sometimes we're dealing with a quadratic function of

[tex]y = ax^2 + bx + c[/tex]

in the second case, we wouldn't write

[tex]y - c = ax^2 + bx[/tex]

because it'd be confusing. That means that when we're looking for y=0 points of a quadratic equation, the way to convey that goal is that we're looking for the point where [tex]ax^2 + bx + c = 0[/tex] - so we preserve the structure of the quadratic function.