i need to solve each equation by using the method of you choice. find exact solutions
B. describe the number and type of roots.
x^2=8x-16 okay im stuck in thiz one too?
2008-12-28 1:49 am
回答 (10)
2008-12-28 1:58 am
✔ 最佳答案
x²= 8x-16, Rearranging the terms we get, x² - 8x +16 = 0x² - 2* x *4 + 4² = 0 is of the form a² - 2ab + b² = 0 =>(a-b)² = 0.
Hence we have, (x-4)²=0 => x-4 = 0
x=4.
The roots are real and equal.
AJM
參考: General Ability in Maths
2008-12-28 9:53 am
x^2 - 8x + 16 = 0
(x-4)(x-4) = 0
x = 4
(x-4)(x-4) = 0
x = 4
2008-12-28 12:50 pm
x² = 8x - 16
x² - 8x = - 16
x² - 4x = - 16 + (- 4)²
x² - 4x = - 16 + 16
(x - 4)² = 0
x - 4 = 0
x = 4
Answer: x = 4
Proof:
4² = 8(4) - 16
16 = 32 - 16
16 = 16
x² - 8x = - 16
x² - 4x = - 16 + (- 4)²
x² - 4x = - 16 + 16
(x - 4)² = 0
x - 4 = 0
x = 4
Answer: x = 4
Proof:
4² = 8(4) - 16
16 = 32 - 16
16 = 16
2008-12-28 9:53 am
x^2-8x+16=0
(x-4)(x-4)=0
x=4
(x-4)(x-4)=0
x=4
2008-12-28 9:53 am
x²-8x+16=0
(x-4)²=0
x-4=0
x=4
Real, rational
(x-4)²=0
x-4=0
x=4
Real, rational
2008-12-28 10:17 am
x^2-8x+16 =
(x-4) * (x-4) = 0
So x = 4
(x-4) * (x-4) = 0
So x = 4
2008-12-28 9:59 am
x^2=8x-16
x^2+8x+16=0
(x+4)^2=0
x=-4
x^2+8x+16=0
(x+4)^2=0
x=-4
2008-12-28 5:51 pm
x^2 = 8x - 16
x^2 - 8x + 16 = 0
x^2 - 4x - 4x + 16 = 0
(x^2 - 4x) - (4x - 16) = 0
x(x - 4) - 4(x - 4) = 0
(x - 4)(x - 4) = 0
x - 4 = 0
x = 4
â´ x = 4
x^2 - 8x + 16 = 0
x^2 - 4x - 4x + 16 = 0
(x^2 - 4x) - (4x - 16) = 0
x(x - 4) - 4(x - 4) = 0
(x - 4)(x - 4) = 0
x - 4 = 0
x = 4
â´ x = 4
2008-12-28 9:56 am
x^2=8x-16
x^2-8x+16=0
(x-4)(x-4)=0
x-4=0 x-4=0
x=4 x=4 answer @(^_^)@
x^2-8x+16=0
(x-4)(x-4)=0
x-4=0 x-4=0
x=4 x=4 answer @(^_^)@
2008-12-28 9:55 am
x^2-8x+16=0
(x-4)(x-4)=0
x1=x2=4
(x-4)(x-4)=0
x1=x2=4
收錄日期: 2021-05-01 11:43:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081227174959AAd5YS5