Answer:
[tex](x+\frac{3}{4})(x+\frac{3}{4})=0[/tex]
Step-by-step explanation:
We are given the quadratic equation [tex]16x^{2}+24x+9=0[/tex]
Now, the roots of the quadratic equation [tex]ax^{2}+bx+c=0[/tex] are given by [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex].
So, from the given equation, we have,
a = 16, b =24 , c = 9.
Substituting the values in [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex], we get,
[tex]x=\frac{-24\pm \sqrt{(24)^{2}-4\times 16\times 9}}{2\times 16}[/tex]
i.e. [tex]x=\frac{-24\pm \sqrt{576-576}}{32}[/tex]
i.e. [tex]x=\frac{-24\pm \sqrt{0}}{32}[/tex]
i.e. [tex]x=\frac{-24}{32}[/tex]
i.e. [tex]x=\frac{-3}{4}[/tex]
Thus, the roots of the equation are [tex]\frac{-3}{4}[/tex] and [tex]\frac{-3}{4}[/tex].
Hence, the factored form of the given expression will be [tex](x+\frac{3}{4})(x+\frac{3}{4})=0[/tex]
