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For what value of p, the equation (3p-1) x^2 + 5x + (2p-3) = 0Will have 0 as one of the roots. Also find other root.

Question

For what value of p, the equation (3p1)x2+5x+(2p3)=0 (3p-1)x^2 + 5x + (2p-3) = 0 will have 0 as one of the roots. Also find the other root.

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Solution

To find the value of p for which the equation (3p-1)x^2 + 5x + (2p-3) = 0 has 0 as one of the roots, we can use the fact that if a quadratic equation has a root of 0, then the constant term of the equation must be 0.

So, we set the equation equal to 0 and solve for p:

(3p-1)x^2 + 5x + (2p-3) = 0

Since we know that one of the roots is 0, we can substitute x = 0 into the equation:

(3p-1)(0)^2 + 5(0) + (2p-3) = 0

Simplifying this equation gives us:

2p - 3 = 0

Adding 3 to both sides of the equation:

2p = 3

Dividing both sides by 2:

p = 3/2

So, the value of p for which the equation has 0 as one of the roots is p = 3/2.

To find the other root, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = (3p-1), b = 5, and c = (2p-3). Substituting these values into the quadratic formula:

x = (-5 ± √(5^2 - 4(3p-1)(2p-3))) / (2(3p-1))

Simplifying this equation will give us the other root of the quadratic equation.

This problem has been solved

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