How do I write a quadratic equation with these roots?

Question

Write a quadratic equation that has the given roots: -6i, 6i

a) x^2+36=0
b) x+36=0
c) x^2-36=0
d) x^2+x-36=0

Answer 1

sum of the roots
=-6i+6i=0
product of the roots
=-6i*6i
=36
Therefore required equation
x^2+0x+36=0
or x^2+36=0
a is correct

Answer 2

If the "roots" to a quadratic are known, take the steps backwards pretending as if the sought equation was just perfectly factored!

Given simplier roots such as 2 and 3, it is slightly easier to see the answer due to the non-complexity of the roots. The same steps can be applied, though. They could be rewritten as x = 2 and x = 3. Continuing backwards, subtract 2 from both sides of one solution, and subtract 3 from both sides of the second solution; this yields x - 2 = 0 and x - 3 = 0. Placing both left sides of the solutions into parenthesis and distributing will yield the unfactored, soughtafter quadratic.

(x - 2)(x -3) = x^2 - 3x - 2x + 6
= x^2 - 5x + 6
And, the final answer should be written x^2 - 5x + 6 = 0.

Now the more complex situation we have. Given the roots 6i and -6i, they can be rewritten as x = 6i and x = -6i. Continuing to work backwards, let's look at one solution, and then the other one. If x = 6i, then 6i can be subtracted from both sides; this yields x - 6i = 0. If x = -6i, then 6i can be added to both sides of this solution; this yields x + 6i = 0.

We know that x + 6i and x - 6i are our perfect factors of the quadratic equation, so if they are put into parenthesis and distributed into one another, they should yield the quadratic equation sought.

(x + 6i) (x - 6i) = x^2 - 6i(x) + 6i(x) - 36i^2
= x^2 + 36

The final answer should be written as x^2 + 36 = 0.

Answer 3

To get roots 9 and -7, we set the quadratic equation like this: (x - 9)(x + 7) = 0 by using fact in case you comprehend the 0 product theory, this could bring about x - 9 = 0 or x + 7 = 0 which then ends up in x = 9 or x = -7 that are the roots given. employing the FOIL technique, (x - 9)(x + 7) = 0 will open as much as be x^2 - 2x - sixty 3 = 0. further for 2.

Answer 4

I'm assuming that this is a multiple choice question.

You need to solve for x to check what the roots are.

a) x^2 + 36 =0

sub 36: x^2 = -36
root: x= +/- rt(-36)
-1 in root: x = +/- rt (-1*36)
solve root: x= +/- 6i
(Okay, so this is the answer, but let's look at why the other ones are not.)

b) x + 36 = 0
sub 36: x = -36

c) x^2 -36 = 0
sub -36: x^2 = 36
root: x = +/- rt(36)
solve root: x = +/- 6

d) x^2 + x - 36 =0
Quad Eq: x = [-b +/- rt (b^2-4ac)]/2a
plug in: x= [-1 +/- rt (1^2 - 4*1*-36)]/2*1
reduce: x = [-1 +/- rt (1+144)]/2
reduce: x= [-1 +/- rt(145)]/2

Answer 5

The answer is c) x^2-36 as this is factorised to (x+6)(x-6) - the rule of the difference of squares. Values of -6 and +6 will render each bracket's value (and thus the value of the equation) ZERO, which is the y-value for any root.

Answer 6

OK
x= -6i, so x+6i=0
x= 6i, so x-6i=0

Now you multiply this two factors, and make that product equal to zero.
(x+6i)(x-6i)= 0
(x+6i)(x-6i)=x²-x(6i)+6i(x)-36i²=0
x²-6xi+6xi-36(-1)=0, since i²= -1

x²+36=0, which is your solution.

Answer 7

x =6i, x= -6i
x-6i=0, x+6i=0
(x-6i)(x+6i)=0
x^2 - (6i)^2=0
x^2 - (-36)=0
x^2+36=0

Answer 8

x² + 36 = 0
x² = - 36
x² = i² 36
x = ± 6i
ANSWER a) is correct.