Question

find the value of k for the following quadratic equaction, so that it should have equal roots kx square +6x+1=0​

Answer 1

EXPLANATION.

→ It should be equal roots → kx² + 6x + 1 = 0.

→ To find the value of k.

→ For real and equal roots → D = 0

→ b² - 4ac = 0

→ (6)² - 4(k)(1) = 0

→ 36 - 4k = 0

→ k = 36/4

→ k = 9.

→ More information.

→ Formula of quadratic equation.

x² - ( a + b) x + ab

→ where, (a + b) = sum of zeroes of quadratic

equation.

→ where, ( ab) = products of zeroes of quadratic

equation.

→ D > 0 Or [ b² - 4ac > 0 ] → roots are real

and unequal.

→ D = 0 Or [ b² - 4ac = 0 ] → roots are real

and equal.

→ D < 0 Or [ b² - 4ac < 0 ] → roots are imaginary.

Answer 2

$\sf kx \bigg(x - 2 \bigg) + 6 = 0 \\ \\ \\ \sf {kx}^{2} - 2kx + 6 = 0 \\ \\ \\ \sf {b}^{2} - 4ac = 0 \\ \\ \\ \sf \bigg( - 2k { \bigg)}^{2} - 4 \bigg(k \bigg) \bigg(c \bigg) = 0 \\ \\ \\ \sf {4k}^{2} = 24k \\ \\ \\ \sf {4k}^{2} - 24k = 0 \\ \\ \\ \sf 4k \bigg( k - 6 \bigg) = 0\\ \\ \\ \\ \\ \sf \color{red}4k = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: k - 6 = 0 \\ \\ \underline{ \boxed{ \sf \color{red}k = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:k = 0}}$