Question

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

Answer 1

Hint: In order to find the value of $ P $ , equate the given equation with the standard quadratic equation, then find the Discriminant of the equation. From the properties of discriminant, check whether it’s negative, positive or equal to zero, then solve the equation to get the value of $ P $ . Discriminant tells us which kind of roots are possible for the equation.

Complete step by step solution:
Since, the highest degree in the equation $ 2P{y^2} - 8y + P = 0 $ is $ 2 $ , that implies it’s a quadratic equation.
Now, given a quadratic equation $ 2P{y^2} - 8y + P = 0 $ , let it be $ f(x) $
 $ f(x) = 2P{y^2} - 8y + P = 0 $
Comparing the equation with the standard Quadratic equation $ a{y^2} + by + c $
 $ a $ becomes $ 2P $
 $ b $ becomes $ - 8 $
And $ c $ becomes $ P $
From the discriminant formula, we know that:
 $ D = {b^2} - 4ac $
Substituting the values of $ a,b,c $ in the above formula in order to calculate the discriminant, we get:
 $ D = {\left( { - 8} \right)^2} - 4\left( {2P} \right)\left( P \right) $
Solving the radicands, we get:
 $ D = 64 - 8{P^2} $ …….(1)
Since, we know that there are three Discriminant rules, which are as follows:
1.If Discriminant is positive that means $ D > 0 $ then there are two real and distinct roots.
2.If Discriminant is zero that means $ D = 0 $ then there are two real and equal roots.
3. If Discriminant is negative that means $ D < 0 $ then there are no real roots.
According to the question, we are given that the equation $ 2P{y^2} - 8y + P = 0 $ has equal roots, which is followed by the second property that gives the information that the Discriminant is equal to zero, $ D = 0 $ .
So, equating the discriminant $ \left( D \right) $ (equation 1) with zero, we get:
 $ D = 64 - 8{P^2} = 0 $
 $ \Rightarrow 64 - 8{P^2} = 0 $
Adding both the sides by $ 8{P^2} $ :
 $ \Rightarrow 64 - 8{P^2} + 8{P^2} = 8{P^2} $
 $ \Rightarrow 64 = 8{P^2} $
Which can be also written as:
 $ \Rightarrow 8{P^2} = 64 $
Dividing both the sides by $ 8 $ , we get:
 $ \Rightarrow \dfrac{{8{P^2}}}{8} = \dfrac{{64}}{8} $
 $ \Rightarrow {P^2} = 8 $
Taking square root both the sides:
 $ \Rightarrow \sqrt {{P^2}} = \sqrt 8 $
Since, we know that $ \sqrt {{x^2}} = x $ that implies $ \sqrt {{P^2}} = P $ and, we also know that $ {\left( { - 2\sqrt 2 } \right)^2} = 8 $ , and $ {\left( {2\sqrt 2 } \right)^2} = 8 $ . Substituting these values in the above equation, we get:
 $ \Rightarrow P = \pm 2\sqrt 2 $
Therefore, the equation $ 2P{y^2} - 8y + P = 0 $ has two equal roots, where $ P = \pm 2\sqrt 2 $ .
So, the correct answer is “ $ P = \pm 2\sqrt 2 $ .”.

Note: Do not forget to compare the given Quadratic equation with the standard one every time.
In order to find the roots of the equation substitute the value of discriminant in the quadratic formula: $ y_1 = \dfrac{{ - b + \sqrt D }}{{2a}} $ and $ y_2 = \dfrac{{ - b - \sqrt D }}{{2a}} $ where $ {y_1} $ , $ {y_2} $ are root to quadratic equation and $ D $ is the discriminant.