1.The sum of a number and its square is 72.Find all possible number.
2.The sum of two number is 30.If the product of the two number is 221, find the two numbers.
3.Two numbers differ by 4.The sum of the squares of the two numbers is 106. Find the two numbers.
4.The area of a triangle is 18cm^2. If the base of triangle is 5cm shorter than its height, find the base of the triangle.
5.The area of a parallelogram is 98cm^2. The base of the parallelogram is twice its height. Find the base of the parallelogram.
Maths 難題
2010-01-02 1:21 am
回答 (2)
2010-01-02 2:41 am
1.Let x be the number.
x2+x=72
x2+x-72=0
(x-8)(x+9)=0
x=8 or x=-9
∴the number=8 or 9.
2.Let x and 30-x be the two number.
x(30-x)=221
30x-x2=221
x2-30x+221=0
(x-17)(x-13)=0
x=17 or x=13
∴the two number =17,13.
3.Let x and x-4 be the two number.
x2+(x-4) 2=106
x2+x2-8x+16=106
2x2-8x-90=0
x2-4x-45=0
(x-9)(x+5)=0
x=9 or x=-5
∴the two number =-5,-9 or 9,5.
4.Let xcm be the base of the triangle.
1/2 x(x+5)=18
x(x+5)=36
x2+5x=36
x2+5x-36=0
(x+9)(x-4)=0
x=-9(rej.) or x=4
∴the base of the triangle=4cm
x2+x=72
x2+x-72=0
(x-8)(x+9)=0
x=8 or x=-9
∴the number=8 or 9.
2.Let x and 30-x be the two number.
x(30-x)=221
30x-x2=221
x2-30x+221=0
(x-17)(x-13)=0
x=17 or x=13
∴the two number =17,13.
3.Let x and x-4 be the two number.
x2+(x-4) 2=106
x2+x2-8x+16=106
2x2-8x-90=0
x2-4x-45=0
(x-9)(x+5)=0
x=9 or x=-5
∴the two number =-5,-9 or 9,5.
4.Let xcm be the base of the triangle.
1/2 x(x+5)=18
x(x+5)=36
x2+5x=36
x2+5x-36=0
(x+9)(x-4)=0
x=-9(rej.) or x=4
∴the base of the triangle=4cm
2010-01-02 2:35 am
1. Let x be the possible number.
x+x^2 = 72
x^2 + x - 72 = 0
(x-8)(x+9) = 0
x=8 or x=-9
The possible number is 8 and -9.
2. Let x be one of the number, then the other number is (30-x).
x(30-x) = 221
30x-x^2 = 221
x^2-30x+221 = 0
(x-17)(x-13) = 0
x = 17 or x = 13
The two number is 13 and 17.
3. Let x be one of the number, then the other number is (x-4)
x^2 + (x-4)^2 = 106
x^2+x^2-8x+16 = 106
2x^2-8x-90 = 0
x^2-4x-45 = 0
(x-9)(x+5) = 0
x = 9 or x = -5
The number is 9 and 5 or -5 and -9
4. Let x cm be the base of the triangle, then the height is (x+5)cm
1/2 (x)(x+5) = 18
x^2+5x-36 = 0
(x-4)(x+9) = 0
x = 4 or x = -9(rejected)
The base of the triangle is 4 cm.
5. Let x cm be the base of the parallelogram, then the height is 2x cm.
x(2x) = 98
2x^2 = 98
x^2 = 49
x = 7 or -7(rejected)
The base of the parallelogram is 7 cm
2010-01-01 18:50:00 補充：
Sorry, the last question, I did it wrongly.
Since the base of the parallelogram is twice its height.
The height should be x/2.
So we have,
x(x/2) = 98
x^2 = 196
x = 14 or -14(rejected)
The base should be 14.
x+x^2 = 72
x^2 + x - 72 = 0
(x-8)(x+9) = 0
x=8 or x=-9
The possible number is 8 and -9.
2. Let x be one of the number, then the other number is (30-x).
x(30-x) = 221
30x-x^2 = 221
x^2-30x+221 = 0
(x-17)(x-13) = 0
x = 17 or x = 13
The two number is 13 and 17.
3. Let x be one of the number, then the other number is (x-4)
x^2 + (x-4)^2 = 106
x^2+x^2-8x+16 = 106
2x^2-8x-90 = 0
x^2-4x-45 = 0
(x-9)(x+5) = 0
x = 9 or x = -5
The number is 9 and 5 or -5 and -9
4. Let x cm be the base of the triangle, then the height is (x+5)cm
1/2 (x)(x+5) = 18
x^2+5x-36 = 0
(x-4)(x+9) = 0
x = 4 or x = -9(rejected)
The base of the triangle is 4 cm.
5. Let x cm be the base of the parallelogram, then the height is 2x cm.
x(2x) = 98
2x^2 = 98
x^2 = 49
x = 7 or -7(rejected)
The base of the parallelogram is 7 cm
2010-01-01 18:50:00 補充：
Sorry, the last question, I did it wrongly.
Since the base of the parallelogram is twice its height.
The height should be x/2.
So we have,
x(x/2) = 98
x^2 = 196
x = 14 or -14(rejected)
The base should be 14.
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