Given:
There is a triangle given as
Required:
We want to find the sutiable form that show that
[tex]x^2+6x-59=0[/tex]
and also complete the square
[tex](x+p)^2-q=0[/tex]
and find the value of p and q
Explanation:
The area of triangle is
[tex]\begin{gathered} \frac{1}{2}(x+1)(x+5)=32 \\ \\ x^2+5x+x+5=64 \\ x^2+6x-59=0 \end{gathered}[/tex]
hence proved for a
Now for second
[tex]\begin{gathered} x^2+6x+9-9-59=0 \\ (x+3)^2-68=0 \end{gathered}[/tex]
now compare with
[tex](x+p)^2-q=0[/tex]
we get
[tex]\begin{gathered} p=3 \\ q=68 \end{gathered}[/tex]
Final answer:
p=3 and q=68
