Question

if x^2 -3x +4 =0 show that x^4 = 3x-20

Answer 1

\[\large{x^2 - 3x + 4 = 0}\]

\[\large{\implies x^2 = 3x-4}\tag{1}\]

Now squaring both sides:

\[\large{(x^2)^2 = (3x-4)^2}\]

\[\large{\implies x^4 = 9x^2 + 16 - 24 x}\tag{2}\]

Now again we use the value of \(\large{x^2}\) from equation 1 in equation 2:

\[\large{x^4 = 9(3x-4) + 16 - 24x}\]
\[\large{\implies x^4 = 27x - 36 + 16 - 24x}\]
\[\large{\implies x^4 = 3x - 20}\tag{3}\]

If you have a doubt in any step feel free to ask :)

Answer 2

wow, thanks so much, and can i ask you another question

Answer 3

Sure :)

Answer 4

Hence, or otherwise, find the equation with roots (a)^4 and (b)^4 if the roots of x^2-3x+4=0 and are a, and b

Answer 5

See an equation with roots \(\alpha\) and \(\beta\) can be written as:

\[\large{x^2 - (\alpha+\beta) + \alpha \beta = 0}\]

Answer 6

yep

Answer 7

If an equation of the format:
\[\large{ax^2+bx+c = 0}\]
has roots \(x_1\) and \(x_2\) then:

\[\large{x_1 x_2 = \cfrac{c}{a}}\]
And
\[\large{x_1+x_2 = \cfrac{-b}{a}}\]

Answer 8

yes i understand this

Answer 9

Now our equation is \(x^2-3x+4 = 0\).

Thus,

\[\large{\alpha \beta = \cfrac{c}{a} = \cfrac{4}{1} = 4}\tag{1}\]

And
\[\large{\alpha + \beta = \cfrac{-b}{a} = \cfrac{-(-3)}{1} = 3}\tag{2}\]

Answer 10

yep

Answer 11

Now the equation with roots \(\large{\alpha^4 , ~\beta^4}\) will be:

\[\large{x^2 - (\alpha^4 + \beta^4) + (\alpha^4\beta^4) = 0}\]
\[\large{\implies x^2 - (\alpha^4 + \beta^4) + (\alpha \beta)^4 = 0}\tag{3}\]

Answer 12

why can you replace a, as a^4, and b as b^4

Answer 13

This is because you are asking for an equation with roots a^4 and b^4

Answer 14

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
See an equation with roots \(\alpha\) and \(\beta\) can be written as:

\[\large{x^2 - (\alpha+\beta) + \alpha \beta = 0}\]
\(\color{blue}{\text{End of Quote}}\)

When roots were a and b the equation was ^^^

When the roots are a^4 and b^4 then equation will be:

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
Now the equation with roots \(\large{\alpha^4 , ~\beta^4}\) will be:

\[\large{x^2 - (\alpha^4 + \beta^4) + (\alpha^4\beta^4) = 0}\]
\[\large{\implies x^2 - (\alpha^4 + \beta^4) + (\alpha \beta)^4 = 0}\tag{3}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 15

oh okay then

Answer 16

so then how do you find a^4 + b^4

Answer 17

Okay then we have:

\[\large{\alpha^4 +\beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2}\]

Also:
\[\large{\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2 \alpha \beta}\]

Answer 18

First you calculate a^2 +b^2 and then calculate a^4 + b^4

Answer 19

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
\[\large{\alpha \beta = \cfrac{c}{a} = \cfrac{4}{1} = 4}\tag{1}\]
\[\large{\alpha + \beta = \cfrac{-b}{a} = \cfrac{-(-3)}{1} = 3}\tag{2}\]
\[\large{\alpha^4 +\beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2}\]
Also:
\[\large{\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2 \alpha \beta}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 20

Can you find out the answer now ?

Answer 21

okay, that was really helpful and thanks again for your help. i got the answer so yeah. :) thanks it means a lot

Answer 22

Your welcome