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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 $\frac{5}{3}$.