write down three quadratic equations with the following root and prove the roots are in fact the values of x for which y is equal to zero

quadratic equations with the following root and prove the roots are in fact the values of x for which y is equal to zero

Equation equals

two real roots

Expert Answers

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The easiest way to do this is to pick two numbers, r and s, and plug them into

y=(x-r)(x-s)

and then distribute.

 

For example, if I want 3 and 4 to be roots of a quadratic:

`y=(x-3)(x-4) = x^2 - 3x - 4x + 12 = x^2 - 7x + 12`

So our first example of a quadratic with two real roots is:

`y=x^2-7x+12`

Now, we check that the values of x for which y is equal to 0 really are 3 and 4:

`(3)^2-7(3)+12=9-21+12=0`

So 3 is really a root, since x=3 means that y=0

Try x=4:

`(4)^2-7(4)+12=16-28+12=0`

So 4 is also a root.

 

Now, we generate a new quadratic.  Say the roots are 2 and -3:

`y=(x-2)(x+3) = x^2 - 2x + 3x - 6 = x^2 + x - 6`

We check that 2 and -3 are really the roots:

`y=(2)^2+2-6 = 4 + 2 - 6 = 0`

`y=(-3)^2+(-3)-6 = 9 - 3 - 6 = 0`

 

Now try the roots 0 and 17:

`y=(x-0)(x-17) = x^2 - 17x`

Check that 0 and 17 really are the roots:

`y=(0)^2-17(0) = 0`

`y=(17)^2-17(17)=289-289 = 0`

 

 

Thus, here are three quadratic equations with two real roots:

`y=x^2-7x+12`

`y=x^2+x-6`

`y=x^2-17x`

 

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