Algebra word problems

Question

One number is 3 larger than another and the sum of their square is 269. Find the number.

 

 

Twice the square of a number equals the sum of 5 times the number and 17. Find the number.

 

 

 

The sum of twice the square of a number and 7 times the number equals 15. What is the number?

Answer 1

x and x+3 are the numbers

x^2+(x+3)^2=269

x^2+x^2+6x+9=269

2x^2+6x+9-269=0

2x^2+6x-260=0

x^2+3x-130=0

(x+13)(x-10)=0

x-10=0

x=10

x+3=13

 

also...

x+13=0

x=-13

x+3=-10

 

2x^2+7x=15

2x^2+7x-15=0

(2x-3)(x+5)=0

x+5=0

x=-5

also...

2x-3=0

2x=3

x=3/2 also works

 

the second problem...

2x^2=5x+17

2x^2-5x-17=0

is the wording wrong ? is it 15 times the number ? try the quadratic formula !

 

 

 

 

 

Answer 2

1) Let's call the first number a and the second number b

    According to the problem, one number is 3 times larger than another. This means that:

            a + 3 = b

 

   The second part of the problem tells you that the sum of their squares is 269, which means:

            a² + b² = 269

 

   We are going to use the second equation to solve the for the numbers, but we have a problem: the second equation has two variables in it. We need to reduce this to one variable. We reduce the number of variables using the first equation. We subtract 3 from both sides to get:

 

            a = b - 3

 

We then replace the "a" in the second equation to get:

 

            (b - 3)² + b² = 269

 

Now, solve for b.

Answer 3

Let x and y be the two unknown numbers:

• One number is 3 larger than the other:  x = y+3
• The sum of their squares is 269: x² + y² = 269

 

Since x = y+3, substitute y+3 in place of x in the second equation above:

 

x² + y² = 269

(y+3)² + y² = 269

y² + 6x + 9 + y² = 269

2y² + 6x - 260 = 0

 

Factors to:

 

(y-10)(2y+26) = 0

y = 10 and -13

 

So:

• When y = 10, x = y+3 = 10+3 = 13
• When y = -13, x = y+3 = -13+3 = -10

 

There are two solutions: {10,13} and {-13,-10}