Question

Find the whole number such that twice its square is 11 more than 21 times yhe original number

Answer 1

Answer:

answer: 11

Explanation:

let X be the whole number

From: "twice its square is 11 more than 21 times the original number"

2x^2 = 21x+11

2x^2 - 21x - 11 = 0

Factoring the left:

(2x+1)(x-11) = 0

.

Setting each term on the left to zero:

(2x+1) = 0

2x = -1

x = -1/2

.

(x-11) = 0

x = 11

.

Solution set:

x = {-1/2, 11}

.

Since the problem asks for a "whole number" we eliminate the -1/2 leaving us with:

x=11

Answer 2

Answer:Let a is the whole number.

Now ,

2×a*2=21a+11

2a*2-21a-11=0

(2a+1)×(a-11)=0

a=11 , -1/2

But the answer is

a=11

Because of -1/2 is not a whole number. Ok

Explanation: