Question 137421
Solve by factoring:
1)
{{{5t^2-10t = 0}}} Factor out 5t from each term:
{{{5t(t-2) = 0}}} Now apply the zero products rule;
{{{5t = 0}}} and/or {{{t-2 = 0}}}
If {{{5t = 0}}} then {{{t = 0}}}
If {{{t-2 = 0}}} then {{{t = 2}}}
2)
{{{x^2-3x-10 = 0}}} Factor the trinomial.
{{{(x+2)(x-5) = 0}}} Apply the zero products rule.
{{{x+2 = 0}}} and/or {{{x-5 = 0}}}
If {{{x+2 = 0}}} then {{{x = -2}}}
If {{{x-5 = 0}}} then {{{x = 5}}}
3)
{{{f(x) = 3x^2+5x-2}}} find: f(-2). Just substitute -2 everywhere there is an x in the function.
{{{f(-2) = 3(-2)^2+5(-2)-2}}} Simplify.
{{{f(-2) = 3(4)+(-10)-2}}}
{{{f(-2) = 12-10-2}}}
{{{f(-2) = 0}}}
Find: f(4)
{{{f(4) = 3(4)^2+5(4)-2}}}
{{{f(4) = 3(16)+20-2}}}
{{{f(4) = 48+20-2}}}
{{{f(4) = 66}}}
4) Solve for x using the quadratic formula:{{{x = (-b+-sqrt(b^2-4ac))/2a}}}
{{{3x^2+19x-14 = 0}}} Here, a = 3, b = 19, and c = -14. Making the appropriate substitutions, we get:
{{{x = (-19+-sqrt((19)^2-4(3)(-14)))/2(3)}}} Simplify this:
{{{x = (-19+-sqrt(361-(-168)))/6}}}
{{{x = (-19+-sqrt(529))/6}}}
{{{x = (-19+23)/6}}} or {{{x = (-19-23)/6}}}
{{{x = 2/3}}} or {{{x = -7}}}