How do you write a quadratic equation in standard form with solutions -2 and 7?

2 Answers
Sep 9, 2016

Answer:

#color(green)(x^2-5x-14=0)#

Explanation:

If #color(red)("(-2))# and #color(blue)(""(7))# are solutions to a quadratic equation #color(green)(f(x)=0)#

then #(x-color(red)(""(-2)))# and #(x-color(blue)(7))# are factors of #color(green)(f(x)#
and we can write
#color(white)("XXX")color(green)(f(x))=(x+2)(x-7)=x^2-5x-14#

Sep 9, 2016

Answer:

#a(x^2-5x-14)=0, a!=0#

Explanation:

Let the reqd. quadr. eqn. be #p(x)=ax^2+bx+c=0, ane0#.

#x=-2" is a soln. "rArr p(-2)=0 rArr 4a-2b+c=0... ... (1)#

Similarly, #p(7)=0 rArr 49a+7b+c=0.............................(2)#

#(2)-(1) rArr 45a+9b=0 rArr b=-5a.............................(3)#

Sub.ing this #b# in #(1), 4a+10a+c=0 rArr c=-14a.........(4)#.

Using #(3) and (4)#, we get the reqd. quadr.

#p(x)=ax^2-5ax-14a=0, or, p(x)=a(x^2-5x-14)=0, a!=0#