Question

Find the values of p for which the quadratic equation has equal roots. Also, find these roots.

Answer 1

The given quadric equation is $\left(2p+1\right)x^2-\left(7p+2\right)x+\left(7p-3\right)=0$, and roots are real and equal.

Then, find the value of p.

Here, $a=2p+1,b=-7p-2\mathrm{and}c=7p-3$.

As we know that $D=b^2-4ac$

Putting the values of $a=2p+1,b=-7p-2\mathrm{and}c=7p-3$.

$$\begin{gathered} D={\left[-\left(7p+2\right)\right]}^2-4\left(2p+1\right)\left(7p-3\right) \\ =(49p^2+28p+4)-4\left(14p^2-6p+7p-3\right) \\ =49p^2+28p+4-56p^2-4p+12 \\ =-7p^2+24p+16 \end{gathered}$$

The given equation will have real and equal roots, if D = 0

Thus, $-7p^2+24p+16=0$
$$\begin{gathered} \Rightarrow 7p^2-24p-16=0 \\ \Rightarrow 7p^2-28p+4p-16=0 \\ \Rightarrow 7p(p-4)+4(p-4)=0 \\ \Rightarrow (7p+4)(p-4)=0 \\ \Rightarrow 7p+4=0\mathrm{or}p-4=0 \\ \Rightarrow p=-\frac{4}{7}\mathrm{or}p=4 \end{gathered}$$

Therefore, the value of p is 4 or $-\frac{4}{7}$.

Now, for p = 4, the equation becomes

$$\begin{gathered} 9x^2-30x+25=0 \\ \Rightarrow 9x^2-15x-15x+25=0 \\ \Rightarrow 3x(3x-5)-5(3x-5)=0 \\ \Rightarrow (3x-5{)}^2=0 \\ \Rightarrow x=\frac{5}{3},\frac{5}{3} \end{gathered}$$

for p = $-\frac{4}{7}$, the equation becomes

$$\begin{gathered} \left(-\frac{8}{7}+1\right)x^2-\left(-4+2\right)x+\left(-4-3\right)=0 \\ \Rightarrow \left(\frac{-8+7}{7}\right)x^2+2x-7=0 \\ \Rightarrow -\frac{1}{7}{x}^2+2x-7=0 \\ \Rightarrow -x^2+14x-49=0 \\ \Rightarrow x^2-14x+49=0 \\ \Rightarrow x^2-7x-7x+49=0 \\ \Rightarrow x(x-7)-7(x-7)=0 \\ \Rightarrow (x-7{)}^2=0 \\ \Rightarrow x=7,7 \end{gathered}$$

Hence, the roots of the equation are $\frac{5}{3}\mathrm{and}7$.