Question

If one root of the equation 8x^2-6x-a-3=0 is the square of the other , then the value of a ?

Answer 1

@jim_thompson5910

Answer 2

\[8x^2-6x-a-3\]

Answer 3

Hint:


one root is r, so the other root is r^2


(x - r)*(x - r^2) = x^2 - r^2*x - r*x + r^3

Answer 4

let the roots be t and t^2
\[\sum~of~roots=t+t^2=\frac{ -b }{ a }=-\frac{ -6 }{ 8 }=\frac{ 3 }{ 4 }\]
product of roots=t*t^2=c/a=\[\frac{ -a-3 }{ 8 }\]
\[t^3=\frac{ -a-3 }{ 8 }\]
\[t+t^2=\frac{ 3 }{ 4 }\]
cubing both sides
\[\left( t+t^2 \right)^3=\left( \frac{ 3 }{ 4 } \right)^3\]
\[t^3+t^6+3t*t^2\left( t+t^2 \right)=\frac{ 27 }{ 64 }\]
\[\frac{ -a-3 }{ 8 }+\left( \frac{ -a-3 }{ 8 } \right)^2+3*\frac{ -a-3 }{ 8 }\left( \frac{ 3 }{ 4 } \right)=\frac{ 27 }{ 64 }\]
multiply by 64
\[-8a-24+a^2+9+6a-18a-54=27,a^2-20a-72=0,a=\frac{ 20\pm \sqrt{400+288} }{ 2 }\]
a=?

Answer 5

If one root is square of other root . question is this

Answer 6

yes
in \[ax^2+bx+c=0\]
sum of roots=-b/a
product of roots=c/a

Answer 7

i don't know but your answer isn't correct

Answer 8

@jim_thompson5910 help

Answer 9

@sshayer you did mistake on first step

Answer 10

(x - r)*(x - r^2) = x^2 - r^2*x - r*x + r^3

8*(x - r)*(x - r^2) = 8*(x^2 - r^2*x - r*x + r^3)

8*(x - r)*(x - r^2) = 8*x^2 + 8*( -r^2 - r)*x + 8*r^3


Compare the x terms

we have -6x and 8*( -r^2 - r)*x

so -6 = 8(-r^2 - r)

solve for r

Answer 11

Note that you'll get two different values for \(r\), and depending on which one you choose, a different value of \(a\). For each value of \(a\), the original equation will have two solutions, and if you do the work correctly, the roots will satisfy the constraint that one is the square of the other. Don't be confused by the fact that the two values of \(r\) you get along the way do not!

Answer 12

^

Answer 13

Hmm this was a fun problem :)

Answer 14

ITS

Answer 15

Figure it out Harry Potter? :)

Answer 16

express whole equation in form
x^2 + px + q = 0 form (here divide throughout by 8)

now if A, B are the roots
then A+B = -p
AB = q
but one root is square of the other
so B = A^2.... not solve by this method