Quite complicated solution. It can be simplified using the formula (a+b)²=a²+2ab+b², for both parts of the equation: - Add 9/4 to both parts of the equation. - The left one becomes (x+3/2)². - The right one becomes: 10000099998 + 9/4 = 10000000000 + 100000 + 1/4 = 100000² + 2•100000•½ + ½²= (100000 + ½)², So x+3/2=±(100000+½), and x=99999, -100002
0. Since the calculator is not allowed, there must be at least 1 way that can extremely simplify the question. 1.x^2 + 3x = x(x+3) = (x+1)(x+2)-2, let u=x+1 2. u(u+1) = 10000099998+2 = 10000100000 = 100000(100000+1), u=100000 is one of root, 3. -( root1 + root2) = coefficient of "u" = 1, u= 100000, -100001 4. x = u-1 = 99999, -100002
I agree, why split 99999 = 9×11111, and then let y = 11111 to require processing 9y, which is then processed by fully (9y)^2 = 81y^2 which is the factorised to (9y)^2, and the final solution is requires 9y? Why not let y = 99999 and use that, simplifying the calculations and reducing the risk of errors?
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Quite complicated solution. It can be simplified using the formula
(a+b)²=a²+2ab+b², for both parts of the equation:
- Add 9/4 to both parts of the equation.
- The left one becomes (x+3/2)².
- The right one becomes:
10000099998 + 9/4 =
10000000000 + 100000 + 1/4 =
100000² + 2•100000•½ + ½²=
(100000 + ½)²,
So x+3/2=±(100000+½),
and x=99999, -100002
0. Since the calculator is not allowed, there must be at least 1 way that can extremely simplify the question.
1.x^2 + 3x = x(x+3) = (x+1)(x+2)-2, let u=x+1
2. u(u+1) = 10000099998+2 = 10000100000 = 100000(100000+1), u=100000 is one of root,
3. -( root1 + root2) = coefficient of "u" = 1, u= 100000, -100001
4. x = u-1 = 99999, -100002
Based on the concept of step 0, 10000099998 is ugly.
We should do something to make 10000099998 +2.
That is the reason I chose this solution.
👍
x^2+3x = (10^5)^2+99999 - 1
x^2+3x = (y+1)^2+y - 1 , where y = 99999
x^2+3 x = y^2+3 y
(x - y)(x+y+3) = 0
x = y , - y - 3
x = 99999 , - 100002
Precisely, why split the 99999 into 9×11111 and have to worry about the coefficient when y=11111?
Simpler calculations reduce risk of errors.
Que lentitud
You’ll never see a question like this on a Cambridge entrance paper.
Way, way too easy.
What was the point of taking 99999=9y?
I agree, why split 99999 = 9×11111, and then let y = 11111 to require processing 9y, which is then processed by fully (9y)^2 = 81y^2 which is the factorised to (9y)^2, and the final solution is requires 9y?
Why not let y = 99999 and use that, simplifying the calculations and reducing the risk of errors?
@cigmorfil4101 I think these people are more interested in making longer videos than teaching