Question

Find the value of p such that the given equations have real and equal roots. (i) px²- (2p+1)x +p =0

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$p=\dfrac{-1}{4}$

Step-by-step explanation:

For any quadratic equation to have real and equal roots, its Determinant must be equal to zero.

Determinant , $\sqrt{b^{2}-4ac}=0$

We have been given the equation:-

$px^{2}-(2p+1)x+p=0$

For it to have equal and real roots:-

$\sqrt{{(2p+1)}^{2}-4(p)(p)}=0 \\ \\ \\{(2p+1)}^{2}-4{p}^{2}=0 \\ \\ \\(2p+1)^{2}=4p^{2} \\ \\ \\4p^{2}+1+4p=4p^{2}\\ \\ \\4p=-1 \\ \\ \\p=\dfrac{-1}{4}$


When 'p' is given the value of $\dfrac{-1}{4}$the equation becomes:-


$\dfrac{-x^{2}}{4} - \left(\dfrac{-1}{2}+1\right)x-\left(\dfrac{-1}{4}\right)=0 \\ \\ \\-\dfrac{x^{2}}{4} -\dfrac{x}{2}-\dfrac{1}{4}=0 \\ \\ \\x^{2}+2x+1=0 \\ \\x^{2}+x+x+1=0 \\ \\x(x+1)+1(x+1)=0 \\ \\(x+1)^{2}=0 \\ \\x = -1$

x has real and equal roots i.e. -1