Question

A quadratic equation whose zeros are -3 and 4, is
(a) \[{{x}^{2}}-x+12=0\]
(b) \[{{x}^{2}}+x+12=0\]
(c) \[\dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6=0\]
(d) \[2{{x}^{2}}+2x-24=0\]

Answer 1

Hint: We solve this question by taking the general form of the equation as \[a{{x}^{2}}+bx+c=0\] and we take the sum of roots \[-\dfrac{b}{a}\] and the product of roots as \[\dfrac{c}{a}\] . Then we find the values of ‘b’ and ‘c’ in terms of ‘a’ and then by substituting the values of ‘b’ and ‘c’ in terms of ‘a’ in the general form we take we will get the required equation.

Complete step-by-step answer:
Let us consider that the general form of quadratic equation as \[a{{x}^{2}}+bx+c=0\] and let us assume that p, q are roots of above equation where \[p=-3\] and \[q=4\]
Now by using the sum of roots concept we can find ‘b’ in terms of ‘a’ as follows
 \[\begin{align}
  & \Rightarrow p+q=-\dfrac{b}{a} \\
 & \Rightarrow -3+4=-\dfrac{b}{a} \\
 & \Rightarrow b=-a \\
\end{align}\]
By using the product of roots concept we can find ‘c’ in terms of ‘a’ as follows
 \[\begin{align}
  & \Rightarrow p.q=\dfrac{c}{a} \\
 & \Rightarrow \left( -3 \right)\left( 4 \right)=\dfrac{c}{a} \\
 & \Rightarrow c=-12a \\
\end{align}\]
By substituting the values of ‘b’ and ‘c’ in the general quadratic equation we will get
 \[\begin{align}
  & \Rightarrow a{{x}^{2}}+bx+c=0 \\
 & \Rightarrow a{{x}^{2}}+\left( -a \right)x+\left( -12a \right)=0 \\
 & \Rightarrow a\left( {{x}^{2}}-x-12 \right)=0 \\
 & \Rightarrow \left( {{x}^{2}}-x-12 \right)=0 \\
\end{align}\]
Here in this equation if we take ‘2’ common out then we will get
 \[\Rightarrow \dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6=0\]
Therefore option (c) is the correct answer.

So, the correct answer is “Option C”.

Note: This question can be solved in other way by using the substitution of roots in the equation given in the options. If any equation in the options satisfies both the roots then that will be the required equation which has ‘-3’ and ‘4’ as roots.
Since we know the answer let us check for that option.
By taking option (c) and substituting \[x=-3\] we get
 \[\begin{align}
  & \Rightarrow LHS=\dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 \\
 & \Rightarrow LHS=\dfrac{{{\left( -3 \right)}^{2}}}{2}-\dfrac{\left( -3 \right)}{2}-6 \\
 & \Rightarrow LHS=0 \\
\end{align}\]
Since LHS=RHS we can say that ‘-3’ satisfies the equation.
Now let us substitute \[x=4\] we get
 \[\begin{align}
  & \Rightarrow LHS=\dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 \\
 & \Rightarrow LHS=\dfrac{{{\left( 4 \right)}^{2}}}{2}-\dfrac{\left( 4 \right)}{2}-6 \\
 & \Rightarrow LHS=0 \\
\end{align}\]
Since LHS=RHS we can say that ‘-3’ satisfies the equation.
Since the equation satisfies both roots we can say that this is the answer.