Question

Find the value of a for which the expression (2a+3)x^2-6x+4-a is a perfect square.

Answer 1

do you know what a "perfect square trinomial" is?

Answer 2

no

Answer 3

how about that

Answer 4

ok... do you know what a perfect square is?

Answer 5

http://www.youtube.com/watch?v=aM1qf2QN8YE

Answer 6

couldn't get it

Answer 7

please help solve

Answer 8

@Hero

Answer 9

\[x ^{2}-\frac{ 6}{2a+3 }x+\frac{ 4-a }{2a+3}=0\]
\[x ^{2}-\frac{ 6 }{2a+3 }x+\left( \frac{ 3 }{2a+3 } \right)^{2}=\frac{ a-4 }{2a+3}+\left( \frac{ 3 }{2a+3}\right)^{2} \]
put R.H.S=0 and find the values of a.

Answer 10

\[\left( x-\frac{ 3 }{2a+3 } \right)^{2}=\frac{ 2a ^{2}+3a-8a-12+9 }{\left( 2a+3 \right)^{2}}\]
\[=\frac{ 2a ^{2}-5a-3 }{\left( 2a+3 \right)^{2} }\]
put R.H.S.=0 and find the values of a.

Answer 11

Another way to do it is discriminants

(Coeff of x)^2 - [4 * (coeff of x^2) * (constant)] = 0 for it to be a perfect square

Answer 12

I knew there was an easier way, just couldn't think of it