If the equation \(2x^2 + kx + 1.25k = 0\) has repeated roots, then the value of \(k\) will be:
\(k = 10 \text{ or } k = 0\)
A quadratic \(ax^2 + bx + c = 0\) has repeated (equal) roots when its discriminant \(b^2 - 4ac = 0\).
Here \(a = 2\), \(b = k\) and \(c = 1.25k\), so discriminant \(= k^2 - 4 \times 2 \times 1.25k = k^2 - 10k\).
Setting \(k^2 - 10k = 0\) gives \(k(k - 10) = 0\).
Therefore \(k = 0\) or \(k = 10\).
Hence, \(k = 10 \text{ or } k = 0\).
Similar Questions
\(x^2 - 2x + 3 = 0\) is a ______.
Find the quadratic equation whose roots are equal to the sum and product of the roots of the quadratic equation \(2x^2 + 4x - 1 = 0\).
One of the roots of the equation \(x^2 - 24x + k = 0\) is \(x = 2\). The other root will be:
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - 2x - 1 = 0\), then which is the equation whose roots are \(\alpha^2\) and \(\beta^2\)?
If the roots of the equation \(x^2 - 2x + 1 = 0\) are \(\alpha\) and \(\beta\), then the equation whose roots are \(\alpha^2\) and \(\beta^2\) will be:
If the difference of the squares of the roots of the equation \(x^2 - 6x + q = 0\) is 24, then find the value of \(q\).
If \(\alpha\) and \(\beta\) are the two roots of the quadratic equation \(6x^2-5x+1=0\), then the sum of their squares will be:
If \(6x^2 + 2kx + k = 0\) has no real solution, find the value of \(k\).
Important Questions from Quadratic Equation
For what values of k, the roots of 9x 2 + 8kx + 16 = 0 are real and equal?
- If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of \(\rm \left( 1+\frac{1}{x} \right) \) is equal to :
If the roots of the equation (q – r)x 2+ (r – p)x + (p – q) = 0 are equal, then which of the following is true?
Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is
The value of m for which one of the root of x 2 - 3x + 2m = 0 is double of the root of x 2 - x + m = 0 is:
