Question

For what value of k, the quadratic equation ky2-4y +1 0, is a perfect square?

Answer 1

Answer:

2

Step-by-step explanation:

ky²-4y+1=0

We know that (a+b)² = a² + b² + 2ab

∴By trial and error method k = 4

Because (2y)²-2(1)(2)y +1²

=(2y-1)²

Answer 2

Answer:

The value of k must be 4 to make the given quadratic equation a perfect square.

Step-by-step explanation:

The quadratic equation is : k·y² - 4·y + 1 = 0

First to make it a complete square : Divide each term by k to make the leading coefficient of x² = 1

$\text{Now the equation becomes : }y^2-4\cdot \frac{y}{k}+\frac{1}{k}=0........(1)\\\\\text{Now, divide the coefficient of y by 2 }-4\cdot \frac{y}{2\times k}=-2\cdot\frac{y}{k}\\\text{in order to make it a complete square.}\\\\\implies \text{The complete square of the above equation is : }(y-\frac{2}{k})^2=0\\\\\text{Now, expanding this. We get }\\\\y^2+\frac{4}{k^2}-\frac{4}{k}\cdot k=0\\\\\text{Now, comparing the constant terms of the above equation with equation 1}\\\\\implies \frac{4}{k^2}=\frac{1}{k}\\\\\implies k = 4$

Hence, The value of k must be 4 to make the given quadratic equation a perfect square.