Question 353837
<pre>
Solve a) x^2+7x-10=0

Write

{{{x^2+7x-10=0}}}

as

{{{1x^2+7x-10=0}}}

and compare it to

{{{ax^2+bx+c=0}}}

and notice that {{{a=1}}}, {{{b=7}}}, and {{{c=-10}}}

Then substitute in the quadratic formula

 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

and get

 {{{x = (-(7) +- sqrt( (7)^2-4*(1)*(-10) ))/(2*(1)) }}}

Then simplify:

 {{{x = (-7 +- sqrt(49+40) )/2 }}}

 {{{x = (-7 +- sqrt(89) )/2 }}}

which is really two answers:

{{{x = (-7 + sqrt(89) )/2 }}} and {{{x = (-7 - sqrt(89) )/2 }}}


b)find the x intercepts of f(x)=x^2+7x-10

 They are the points:

{{{P((-7 - sqrt(89) )/2,0) }}} and {{{Q((-7 + sqrt(89) )/2,0) }}}

They are the points marked P and Q below on the graph of {{{"f(x)"=x^2+7x-10}}} 


{{{drawing(400,800,-10,5,-25,5,
locate(-7.9,.7,P), locate(1.7,.7,Q),
graph(400,800,-10,4,-25,5,x^2+7x-10) )}}}

If you get the decimal approximations with a calculator,
you'll find that P is approximately the point P(-8.2,0) and
Q is approximately the point Q(1.2,0)

Edwin</pre>