MHB Find a & b for all real values of x

[mathlearn]

Find a & b such that $x^2-\frac{9}{16}=\left(x-a\right)\left(x-b\right)$ for all real values of x;

Looks like the LHS is a difference of two squares, how to find a & b ?

Any Ideas? (Thinking)

 

[MarkFL]

You are correct, the LHS is the difference of squares, and can be written as:

$$x^2-\left(\frac{3}{4}\right)^2$$

Now, how would you factor that difference of squares?

 

[mathlearn]

You are correct, the LHS is the difference of squares, and can be written as:

$$x^2-\left(\frac{3}{4}\right)^2$$

Now, how would you factor that difference of squares?

The factors of two different squares would be ,

$a^2-b^2= (a+b)(a-b)$,

In this case it should be (x-a) & (x-b)
-a & -b are two different variables, and both of them are negative.

So what should be done? (Thinking)

 

[MarkFL]

$$x^2-\left(\frac{3}{4}\right)^2=\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\tag{1}$$

Now, observing that:

$$(c+d)=(c-(-d))$$

Can you write the factorization in (1) in the form you need?

 

[mathlearn]

$$x^2-\left(\frac{3}{4}\right)^2=\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\tag{1}$$

Now, observing that:

$$(c+d)=(c-(-d))$$

Can you write the factorization in (1) in the form you need?

(Speechless) My apologies, Can you demonstrate ?

 

[MarkFL]

$$x^2-\left(\frac{3}{4}\right)^2=\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)=\left(x-\left(-\frac{3}{4}\right)\right)\left(x-\frac{3}{4}\right)$$

So, given the commutative property of multiplication, we may state:

$$(a,b)=\left(\pm\frac{3}{4},\mp\frac{3}{4}\right)$$

 

[mathlearn]

$$x^2-\left(\frac{3}{4}\right)^2=\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)=\left(x-\left(-\frac{3}{4}\right)\right)\left(x-\frac{3}{4}\right)$$

So, given the commutative property of multiplication, we may state:

$$(a,b)=\left(\pm\frac{3}{4},\mp\frac{3}{4}\right)$$

Many thanks (Smile)