Question

If one root of equation \[{{x}^{2}}+ax+12=0\] is 4 while the equation \[{{x}^{2}}+ax+b=0\] has equal roots, then the values of b is
(a) \[\dfrac{4}{49}\]
(b) \[\dfrac{49}{4}\]
(c) \[\dfrac{7}{4}\]
(d) \[\dfrac{4}{7}\]

Answer 1

Hint: Put x = 4 in equation 1. Thus get the value of a. As they have equal roots, we can say that \[D={{b}^{2}}-4ac\] for equation (2). D = 0, thus get the value of b in equation (2) but substituting required values in D.

Complete step-by-step answer:

We have been given one equation as, \[{{x}^{2}}+ax+12=0-(1)\].
Now the root of this equation is given as x = 4.
The equation (1) is of the form of a quadratic equation whose general form is \[a{{x}^{2}}+bx+c=0\].
Now let us substitute x = 4 in equation (1).
At x = 4, \[{{x}^{2}}+ax+12=0\]
\[\begin{align}
  & {{\left( 4 \right)}^{2}}+4a+12=0 \\
 & 16+4a+12=0 \\
 & 4a=-\left( 16+12 \right) \\
 & a=\dfrac{-28}{4}=-7 \\
\end{align}\]
Thus we got the value of a = - 7.
Now we have been given a second equation, \[{{x}^{2}}+ax+b=0-(2)\].
It is said that \[{{x}^{2}}+ax+12=0\] and \[{{x}^{2}}+ax+b=0\] has equal roots, which means that the root of \[{{x}^{2}}+ax+b=0\] is also x = 4.
\[{{x}^{2}}+ax+b=0-(2)\]
We know that, \[D={{b}^{2}}-4ac\], as roots are the same, Discriminant (D) = 0.
\[\therefore {{b}^{2}}-4ac=0-(3)\]
From (2) we know that a = 1, b = a, c = b, from comparing equation (2) the general equation \[a{{x}^{2}}+bx+c=0\].
a = 1, b = a = -7, c =b.
\[\begin{align}
  & {{b}^{2}}-4ac=0 \\
 & {{\left( -7 \right)}^{2}}-4\times 1\times b=0 \\
 & 49-4b=0 \\
 & \Rightarrow 49=4b \\
 & \therefore b=\dfrac{49}{4} \\
\end{align}\]
Thus we got the value of \[b=\dfrac{49}{4}\].
\[\therefore \] Option (b) is the correct answer.

Note: The number \[D={{b}^{2}}-4ac\] is called discriminant.
If D < 0, then the quadratic equation has no real solutions
If D = 0, then the quadratic equation has 1 solution, \[x=\dfrac{-b}{2a}\].
If D > 0, then the quadratic equation has 2 distinct solution