Here's my method, my usual method that has very quickly worked for me with all these "entrance" problems, never mind Harvard, Stanford, MIT etc etc., which are irrational. Let's see, we need a fourth root, so let's make it simple. Sixteen has a simple fourth root, and....that works! So X = 16. It is immaterial to me whether in about fifteen seconds I have demonstrated aptitude. I like this method, because it leaves a lot of time left over for something else.
Why do you have to reject irrational solutions? One of them it is obvious that it is extraneous but the other one does seem pretty close to correct when you plug it back in.
_√x + √(⁴√x + 7) = 7_
Let _a = ⁴√x, b = √(a + 7)_
⇒ _b² - a = 7_ ... ①
We have: _a² + b = 7_ ... ② (the given equation)
Subtract ① from ②:
_(a² - b²) + (a + b) = 0_
⇒ _(a + b)(a - b + 1) = 0_
_a + b > 0 ∴ a - b + 1 = 0_
⇒ _b = a + 1_
Sub into equation ②:
_a² + a - 6 = 0_
⇒ _(a - 2)(a + 3) = 0_
_a ≥ 0 ∴ a = 2_
⇒ _⁴√x = 2_
⇒ *_x = 2⁴ = 16_*
Nice substitution. Straight to the point👌✅
Harvard University Entrance Algebra tricks: √x + √(⁴√x + 7) = 7; x =?
Let: y = ⁴√x > 0, x = y⁴; √x + √(⁴√x + 7) = y² + √(y + 7) = 7, √(y + 7) = 7 - y²
y + 7 = (7 - y²)², y⁴ - 14y² - y + 42 = 0, (y⁴ - 4y²) - (10y² + y - 42) = 0
y²(y - 2)(y + 2) - (10y + 21)(y - 2) = (y - 2)(y³ + 2y² - 10y - 21) = 0
y³ + 2y² - 10y - 21 = (y³ + 2y² - 3y) - (7y + 21) = (y + 3)(y² - y) - 7(y + 3)
= (y + 3)(y² - y - 7); y⁴ - 14y - y + 42 = (y - 2)(y + 3)(y² - y - 7) = 0, y + 3 > 0
y² - y - 7 = (y² - 7) - y = - [√(y + 7) + y] < 0; y - 2 = 0, y = 2 = ⁴√x, x = 2⁴ = 16
Answer check:
x = 16: √x + √(⁴√x + 7) = √16 + √(⁴√16 + 7) = 4 + √9 = 4 + 3 = 7; Confirmed
Final answer:
x = 16
Here's my method, my usual method that has very quickly worked for me with all these "entrance" problems, never mind Harvard, Stanford, MIT etc etc., which are irrational. Let's see, we need a fourth root, so let's make it simple. Sixteen has a simple fourth root, and....that works! So X = 16. It is immaterial to me whether in about fifteen seconds I have demonstrated aptitude. I like this method, because it leaves a lot of time left over for something else.
Your approach is nonstandard in exam setting.
Great method - check against LHS - square root of 16 is 4.....looking good! Plug it all in - VOILA!
@@superacademy247No points for efficiency?
1^2 = 1
1^4 =1
2^2=4
2^4=16
3^2=9
3^4=81
if x is 81
9+ value > 7
try x as 16 and 1
Answer is 16
Why do you have to reject irrational solutions? One of them it is obvious that it is extraneous but the other one does seem pretty close to correct when you plug it back in.
It's obeying the condition imposed on the question
{x+x ➖}=x^2 {x+x+x+x+x ➖ x ➖ x ➖ x}=x^4 {x^2+x^4}=x^6 {7+7 ➖ }{7+7 ➖}={14+14}=28 {x^6+28}=28x^6 7^4x^6 7^1^4x^6 1^1^4x^3^2 2^2x^3^2 1^2x3^1 2x^3 (x ➖ 3x+2).