Question

If roots of the quadratic equation \[3{y^2} + ky + 12 = 0\] are real and equal, then find K.

Answer 1

Hint: An equation where the highest exponent of variable is of degree \[2\]is called a quadratic equation. General form of the quadratic equation is \[a{x^2} + bx + c = 0\]and its roots can be calculated by the using formula of discriminant.
Discriminant:
\[(D)\, = \,{b^2} - 4ac\] and it has \[3\]condition.
(1). If value of\[D > 0\], Roots are equal and distinct and given as :
\[\alpha = \dfrac{{ - b + \sqrt D }}{{2a}}\], \[\beta = \dfrac{{ - b - \sqrt D }}{{2a}}\]
(2). If the value of\[D = 0\], Roots are real and equal and given as: \[\left( {\dfrac{{ - b}}{{2a}}} \right)\].
(3). If the value of\[D < 0\], Roots does not exist i.e. imaginary.
Here, in this question\[D = 0\].

Complete step-by-step solution:
Given: \[3{y^2} + ky + 12 = 0\].
On comparing with general equation \[a{x^2} + bx + c = 0\], we have:
Here, \[a = 3\], \[b = k\], \[c = 12\]
We are given that the roots of the quadratic equation are real and equal. Hence discriminant is zero.
\[\therefore \,\,D = 0\] i.e.
\[ \Rightarrow {b^2} - 4ac = 0\].
On substituting the values of \[a = 3\], \[b = k\]and \[c = 12\], we have:
\[ \Rightarrow {k^2} - 4 \times 3 \times 12 = 0\]
\[ \Rightarrow {k^2} - 144 = 0\]
\[ \Rightarrow {k^2} = 144\]
On taking square root on both sides of the equation, we get:
\[ \Rightarrow k = \sqrt {144} \]
\[ \Rightarrow k = \pm 12\]

Hence the value of k is either 12 or -12.

Note:Roots of quadratic equation can also be calculated by splitting the middle term and completing the square method.
When we take the square root of the equation we get the positive as well as the negative value because the square of both are the same.
If roots are real and distinct, we use condition\[D > 0\].