Find the possible values of $p$ for which the equation has coincident roots.

Question

Find the possible values of $p$ for which the equation $(2p+3)x^2+(4p-14)x+16p+1=0$ has coincident roots.

Coincident roots means 'equal roots'. For equal roots, we should use: $b^2-4ac=0$

$(4p-14)^2-4(2p+3)(16p+1)=0$

After solving this, I get: $-112p^2-536p+184=0$

I solved the above equation by this formula: $\frac{-b\pm \sqrt {b^2-4ac}}{2a}$

Is this right? Maybe there is some mistake but I can't find it. My answers are wrong. I got $-5.11$, $0.32$, my book says: $\frac{1}{2}$, $-3\frac{2}{7}$. Help. :(

Answer 1

You should have $-112p^2 -312p + 184=0$

Answer 2

If you expand the equation $$(4p-14)^2 - 4(2p+3)(16p+1)=0$$ you'll get $$16p² -112p +196 -128p² -200p -12 = 0$$

then, you'll get $$-112p² -312p +184$$

using the Quadratic Formula, you'll get:

$$\dfrac{1}{2}\quad\text{or}\quad-3\frac{2}{7}$$

Answer 3

If it has coincident roots then $b^2-4ac = 0$ $$(4p-1)^2 - 4(2p+3)(16p+1) = 0$$ Expand and you get $$14p^2 + 39p - 23 = 0$$ $$(2p-1)(7p+23) = 0$$ $$P = 1/2\quad\text{or}\quad P = -23/7$$