Question

Can you guys help me please T.T ...
Given that the quadratic equation x^2-2x-5=0 has 2 different roots, and a second quadratic equation has 2 roots, each of which is 2 less than the corresponding root of the given quadratic equation. If the second quadratic equation is x^2+ax+b=0, find the value of a and b.

Answer 1

wow. i think you actually have to find the roots. we can do that fairly easily

Answer 2

root are 2 and -5

Answer 3

\[x^2-2x=5\]
\[(x-1)^2=5+1=6\]
\[x--=\pm\sqrt{6}\]
\[x=1\pm\sqrt{6}\]

Answer 4

so the two roots are
\[1-\sqrt{6}\] and
\[1-\sqrt{6}\] subtract 2 from each get
\[-1-\sqrt{6}\] and
\[-1+\sqrt{6}\]

Answer 5

now multiply out and see what you get in the equation
\[(x-(-1-\sqrt{6}))(x-(-1+\sqrt{6})\] which looks hard but in fact is very easy

Answer 6

*expression

Answer 7

first
\[x^2\] last
\[(-1-\sqrt{6})(-1+\sqrt{6})=1-6=-5\] hmm just like the orignal

Answer 8

"outer, inner"
\[(1-\sqrt{6})x+(1+\sqrt{6})x=2x\]

Answer 9

so equation is
\[x^2+2x-5=0\]

Answer 10

they want you to find the a and b value

Answer 11

You have them,\[ax ^{2}+2x-5=0\]
a=1
b=2