# Find the values of p for which the quadratic equation $(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$ has equal roots. Also, find these roots.

Given:

Given quadratic equation is $(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$.

To do:

We have to find the value of p for which the given quadratic equation has equal roots.

Solution:

$(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2p+1, b=-(7p+2)$ and $c=7p-3$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[-(7p+2)]^2-4(2p+1)(7p-3)$

$D=(7p+2)^2-(8p+4)(7p-3)$

$D=(7p)^2+2(7p)(2)+(2)^2-8p(7p)-8p(-3)-4(7p)-4(-3)$

$D=49p^2+28p+4-56p^2+24p-28p+12$

$D=-7p^2+24p+16$

The given quadratic equation has equal roots if $D=0$.

Therefore,

$-7p^2+24p+16=0$

$7p^2-24p-16=0$

$7p^2-28p+4p-16=0$

$7p(p-4)+4(p-4)=0$

$(7p+4)(p-4)=0$

$7p+4=0$ or $p-4=0$

$7p=-4$ or $p=4$

$p=\frac{-4}{7}$ or $p=4$

The values of p are $\frac{-4}{7}$ and $4$.

For $p = \frac{-4}{7}$,

$[2(\frac{-4}{7}) + 1]x^2 - [7(\frac{-4}{7}) + 2]x + [7(\frac{-4}{7}) - 3] = 0$

$(\frac{-8+7}{7})x^2 - (-4+2)x + (-4-3) = 0$

$(\frac{-1}{7})x^2+2x - 7 = 0$

$x^2+(-7)(2)x+(-7)(-7)=0$     (Multiplying both sides by $-7$)

$x^2-14x+49=0$

$(x-7)^2 = 0$

$x-7=0$

$x=7$

Therefore, for $p=\frac{-4}{7}$ the roots of the given quadratic equation are $7$ and $7$.

For $p = 4$,

$(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$

$[2(4)+1]x^2 -[7(4)+2]x + [7(4)-3] = 0$

$9x^2-30x+25=0$

$(3x)^2-2(3x)(5)+(5)^2=0$

$(3x - 5)^2 = 0$

$3x-5=0$

$3x=5$

$x=\frac{5}{3}$

Therefore, for $p=4$ the roots of the given quadratic equation are $\frac{5}{3}$ and $\frac{5}{3}$.

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