Prove that if one solution for a quadratic equation of theform x2 + bx + c = 0 is rational (where b and c are rational),then the other solution is also rational. (Use the factthat if the solutions of the equation are r and s, thenx2 + bx + c = (x − r )(x − s).)

The other solution of the given equation x² + bx + c = 0   is also rational number.

Step-by-step explanation:

Here, given: ax² + bx + c = 0 is a quadratic equation

Also, one solution  (r)  of equation is RATIONAL.

To show: The other solution (s)  is also RATIONAL

Now, here: as x² + bx + c = 0

Since r and s are the two given solutions, the given equation can be factorized as:

x² + bx + c =  (x -r) (x - s)

Simplifying LHS, we get:

(x -r) (x - s) = x x - r (x) - s (x) +  (r)(s)

                 =  x² + x(-r - s) +  rs

or, x² + bx + c = x² + x(-r - s) +  rs

Comparing the related terms, we get:

b =  (-r - s)    

⇒  b +  s = - r

or, s  = -r - b

Now, given : r = Rational  and the negative of a rational is also rational.

⇒  -r is also rational

Also, difference of two rational number is also rational.

⇒ -r - b is also rational

⇒ s is a RATIONAL NUMBER

Hence, the other solution of the given equation x² + bx + c = 0   is also rational number.

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

ratio is 2 : 9

the answer is three

13.5 : 3