Evaluating Complex Equation without a Calculator

In summary, the conversation discusses evaluating a complex mathematical expression without the use of a calculator. The equation involves multiple square roots and fractions, making it difficult to solve by hand. However, the individual involved suggests using a calculator for the in between steps, but is declined by another person. The original person is thanked for their effort and the exchange ends with a compliment.
  • #1
anemone
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Without the help of calculator, evaluate \(\displaystyle \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}\).
 
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  • #2
Can I use a calculator for just the in between steps? (Kiss)

-Dan
 
  • #3
topsquark said:
(Kiss)

-Dan
Thanks, I'm flattered, but no, thanks.:p

Report for MHB in the beginning of 2016 (Week #4):

When it comes to entertaining and being funny, Dan really takes the cake. Haha!
 
  • #4
topsquark said:
Can I use a calculator for just the in between steps? (Kiss)

-Dan

Since anemone declined the KISS...here you go:

giphy.gif
 
  • #5
let
$x= (-\sqrt{6}+\sqrt{7}+\sqrt{8})$
$y= (\sqrt{6}-\sqrt{7}+\sqrt{8})$
$z=(\sqrt{6}+\sqrt{7}-\sqrt{8})$
so we get
$x-y= - 2(\sqrt{6} - \sqrt{7}) $ , $x-z= - 2(\sqrt{6} - \sqrt{8})$,$y-z = 2(\sqrt{8} - \sqrt{7})$
and $x + y= 2\sqrt{8} $ , $x + z= 2\sqrt{7}$,$y+ z = 2\sqrt{6}$
above sum reduces to
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
now
$x^4(y-z) + y^4(z-x) + z^4(x-y)$
= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$
= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$
= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$
= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$
= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$
=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$
= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$
= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$
= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$
= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$
= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$
so the given expression
= $x^2 + y^2 +z^2 + xy + yz+ xz$
= $\frac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2) = \frac{1}{2}(4 * (8+ 6 + 7)) = 42$
 
  • #6
kaliprasad said:
let
$x= (-\sqrt{6}+\sqrt{7}+\sqrt{8})$
$y= (\sqrt{6}-\sqrt{7}+\sqrt{8})$
$z=(\sqrt{6}+\sqrt{7}-\sqrt{8})$
so we get
$x-y= - 2(\sqrt{6} - \sqrt{7}) $ , $x-z= - 2(\sqrt{6} - \sqrt{8})$,$y-z = 2(\sqrt{8} - \sqrt{7})$
and $x + y= 2\sqrt{8} $ , $x + z= 2\sqrt{7}$,$y+ z = 2\sqrt{6}$
above sum reduces to
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
now
$x^4(y-z) + y^4(z-x) + z^4(x-y)$
= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$
= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$
= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$
= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$
= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$
=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$
= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$
= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$
= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$
= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$
= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$
so the given expression
= $x^2 + y^2 +z^2 + xy + yz+ xz$
= $\frac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2) = \frac{1}{2}(4 * (8+ 6 + 7)) = 42$

Aww...that was exactly how I approached the problem as well! :cool:Thanks kaliprasad for participating!
 
  • #7
kaliprasad said:
let
$x= (-\sqrt{6}+\sqrt{7}+\sqrt{8})$
$y= (\sqrt{6}-\sqrt{7}+\sqrt{8})$
$z=(\sqrt{6}+\sqrt{7}-\sqrt{8})$
so we get
$x-y= - 2(\sqrt{6} - \sqrt{7}) $ , $x-z= - 2(\sqrt{6} - \sqrt{8})$,$y-z = 2(\sqrt{8} - \sqrt{7})$
and $x + y= 2\sqrt{8} $ , $x + z= 2\sqrt{7}$,$y+ z = 2\sqrt{6}$
above sum reduces to
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
now
$x^4(y-z) + y^4(z-x) + z^4(x-y)$
= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$
= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$
= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$
= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$
= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$
=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$
= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$
= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$
= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$
= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$
= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$
so the given expression
= $x^2 + y^2 +z^2 + xy + yz+ xz$
= $\frac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2) = \frac{1}{2}(4 * (8+ 6 + 7)) = 42$
Wow. For both figuring this out and for typing the whole thing! (Bow)

-Dan
 

FAQ: Evaluating Complex Equation without a Calculator

How do I evaluate a complex equation without a calculator?

Evaluating a complex equation without a calculator involves breaking down the equation into smaller, simpler parts and using algebraic manipulation to simplify the equations. This can also involve using basic arithmetic operations and known mathematical properties to solve the equation step by step.

What are some strategies for evaluating complex equations without a calculator?

Some strategies for evaluating complex equations without a calculator include simplifying the equation by factoring, using the distributive property, or combining like terms. Breaking down the equation into smaller parts, using inverse operations, and substituting known values can also be helpful.

Can I use estimation when evaluating complex equations without a calculator?

Yes, estimation can be a useful tool when evaluating complex equations without a calculator. Estimation involves rounding numbers to the nearest whole number or decimal place and using mental math to solve the equation. This can give you a rough idea of the solution and can be helpful in checking your work.

How do I know if my answer is correct when evaluating complex equations without a calculator?

One way to check the accuracy of your answer when evaluating complex equations without a calculator is to use estimation or plug your solution back into the original equation to see if it satisfies the equation. Another option is to use online calculators or ask a teacher or tutor for verification.

Are there any tips for becoming more proficient in evaluating complex equations without a calculator?

Practice is key to becoming more proficient in evaluating complex equations without a calculator. Familiarizing yourself with basic algebraic principles, developing mental math skills, and working on various types of equations can also help improve your proficiency. Additionally, seeking guidance from a teacher or tutor can be beneficial in mastering this skill.

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