Solving a quadratic with two terms: 9X^2 - 4 = 0

[MathsFormula]

Hello. The quadratic is 9X² - 4 = 0

I need to factorised in this exercise but there are no common factors to make brackets.

So tried X² =4/9

X² = 2/3

Don't know what to do next.

Answer = -2/3 or 2/3

Thanks

 

[Deleted member 4993]

Hello. The quadratic is 9X² - 4 = 0

I need to factorised in this exercise but there are no common factors to make brackets.

So tried X² =4/9 → X² - (2/3)² = 0

X² = 2/3

Don't know what to do next.

Answer = -2/3 or 2/3

Thanks

You need to use:

a² - b² = (a - b) * ( a + b).

In this case:

a= (X) and

b = 2/3

 

[HallsofIvy]

Hello. The quadratic is 9X² - 4 = 0

I need to factorised in this exercise but there are no common factors to make brackets.

So tried X² =4/9

X² = 2/3

Don't know what to do next.

Answer = -2/3 or 2/3

Thanks

In general, if x= p and x= q are roots of a quadratic equation, then the quadratic can be factored as a(x- p)(x- q) where "a" is the "leading coefficient", the coefficient of \(\displaystyle x^2\).

In this case, since x= 2/3 and x= -2/3 are the roots and the leading coefficient is 9, 9x^2- 4= 9(x- 2/3)(x+ 2/3).

 

[MathsFormula]

Thanks alot. I've more or less understood. Will get back to you all if I can't get a deeper understanding once I've gone over it a few times

 

[Otis]

9X² - 4 = 0

I need to factorised in this exercise

Actually, you don't, unless you were instructed to "solve by factoring".

The method that you tried works, but you forgot about the negative root (and you made a typo).

So tried X² = 4/9

X² = 2/3

That exponent in red is a typo, yes?

When you take the square root of each side of X^2=4/9, the result should be:

|X| = 2/3

Now, to write the solution without absolute-value symbols, we need to state both the positive and negative root:

X = 2/3 or X = -2/3

 

[MathsFormula]

Actually, you don't, unless you were instructed to "solve by factoring".

The method that you tried works, but you forgot about the negative root (and you made a typo).




That exponent in red is a typo, yes?

Unfortunately NO. I just cancelled down the 4/9

 

[MathsFormula]

You need to use:

a² - b² = (a - b) * ( a + b).

In this case:

a= (X) and

b = 2/3

Thanks UNDERSTOOD

 

[stapel]

The quadratic is 9X² - 4 = 0

I need to factorise in this exercise but there are no common factors to make brackets.

Recall the rule for factoring differences of squares:

. . . . .\(\displaystyle a^2\, -\, b^2\, =\, (a\, -\, b)\,(a\, +\, b)\)

In your case, a = 3x and b = 2.

 

[MathsFormula]

Recall the rule for factoring differences of squares:

. . . . .\(\displaystyle a^2\, -\, b^2\, =\, (a\, -\, b)\,(a\, +\, b)\)

In your case, a = 3x and b = 2.

I see it now. Thanks

 

[lookagain]

Unfortunately NO. I just cancelled down the 4/9

4/9 is already in lowest terms. \(\displaystyle \ \ \ \dfrac{4}{9} \ \ne \ \dfrac{2}{3}\)

 

[MathsFormula]

4/9 is already in lowest terms. \(\displaystyle \ \ \ \dfrac{4}{9} \ \ne \ \dfrac{2}{3}\)

That was my foolish mistake