Answer:
(1). [tex]x^2+24x+144,\ x^2+2x+1[/tex]
(2). [tex]\frac{25}{4}[/tex]
(3). [tex]x^2+x+\frac{1}{4}=12+\frac{1}{4}[/tex]
Step-by-step explanation :
Perfect square trinomials :
An equation having three term and which is also a perfect square.
[tex]ax^2+bx+c\ is\ a\ perfect\ square\ if\ b^2-4ac=0.[/tex]
(1).
First Equation :
[tex]x^2+24x+144\\\\a=1,\ b=24\ c=144\\\\b^2-4ac=(24)^2-4\times144\\\\b^2-4ac=576-576\\\\b^2-4ac=0\\\\[/tex]
Hence it is a perfect square.
Second Equation :
[tex]x^2+5x+\frac{25}{6}\\\\a=1,\ b=5,\ c=\frac{25}{6}\\\\b^2-4ac=5^2-4\times1\times\frac{25}{6}\\\\b^2-4ac=25-\frac{100}{6}=\frac{25}{3}[/tex]
It is not a perfect square.
Third Equation :
[tex]x^2+2x+1\\\\a=1,\ b=2,\ c=1\\\\b^2-4ac=2^2-4\times1\times1\\\\b^2-4ac=4-4\\\\b^2-4ac=0[/tex]
It is a perfect square.
Forth Equation :
[tex]x^2+3x+2.62\\\\a=1,\ b=3,\ c=2.62\\\\b^2-4ac=3^2-4\times1\times2.62\\\\b^2-4ac=9-10.48\\\\b^2-4ac=-1.48[/tex]
It is not a perfect square.
(2).
[tex]x^2+5x+n\\\\a=1,\ b=5,\ c=n\\\\For\ perfect\ square\\\\b^2-4ac=0\\\\5^2-4\times1\times n=0\\\\25-4n=0\\\\4n=25\\\\n=\frac{25}{4}[/tex]
(3).
[tex]x^2+x=12[/tex]
To solve it make [tex]L.H.S.[/tex] perfect square.
[tex]L.H.S.\ =x^2+x+c\\\\a=1,\ b=1\\\\b^2-4ac=0\\\\1-4c=0\\\\c=\frac{1}{4}[/tex]
So add [tex]\frac{1}{4}[/tex] both sides to make it perfect square
[tex]x^2+x+\frac{1}{4}=12+\frac{1}{4}[/tex]
