Find the value of $P$ in the equation $2 P y^{2} - 8 y + P = 0$ , so that equation has equal roots.

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Solution

Given equation: $2 P y^{2} - 8 y + P = 0$
On comparing by the general equation
$a x^{2} + b x + c = 0$
We get, $a = 2 P, b = - 8, c = P$
If the equation has equal roots, then
$b^{2} - 4 a c = 0$
$\Rightarrow (- 8)^{2 - 4} \times 2 P \times P = 0$
$\Rightarrow 64 - 8 P^{2} = 0$
$\Rightarrow 8 P^{2} = 0$
$\Rightarrow P^{2} = \frac{64}{8}$
$\Rightarrow P^{2} = 8$
$\Rightarrow P = \pm 2 \sqrt{2}$ ,
Thus for the value $P = \pm 2 \sqrt{2}$ , the equation $2 P y^{2} - 8 y + P = 0$ has equal roots.

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Find the value of P in the equation 2Py2−8y+P=0, so that equation has equal roots.

Given equation: 2Py2−8y+P=0On comparing by the general equationax2+bx+c=0We get, a=2P,b=−8,c=PIf the equation has equal roots, thenb2−4ac=0⇒(−8)2−4×2P×P=0⇒64−8P2=0⇒8P2=0⇒P2=648⇒P2=8⇒P=±2√2,Thus for the value P=±2√2, the equation 2Py2−8y+P=0 has equal roots.

Find the value of $P$ in the equation $2 P y^{2} - 8 y + P = 0$ , so that equation has equal roots.