Question

If the discriminant of \[3{x^2} + 2x + a = 0\] is double the discriminant of \[{x^2} - 4x + 2 = 0\], then find the value of \[a\].

Answer 1

Hint: Here, we need to find the value of \[a\]. We will use the formula for discriminant to find the discriminants of the two given equations. Then, we will use the given information from an equation and solve it to find the value of \[a\].

Formula Used:
 We will use the formula of the discriminant of a quadratic equation of the form \[a{x^2} + bx + c = 0\] is given by \[D = {b^2} - 4ac\].

Complete step-by-step answer:
First, we will find the discriminant of a quadratic equation \[3{x^2} + 2x + a = 0\].
Comparing the equations \[3{x^2} + 2x + a = 0\] and \[a{x^2} + bx + c = 0\], we get
\[a = 3\], \[b = 2\], and \[c = a\]
Substituting \[a = 3\], \[b = 2\], and \[c = a\] in the formula \[D = {b^2} - 4ac\], we get
\[ \Rightarrow D = {2^2} - 4\left( 3 \right)\left( a \right)\]
Applying the exponent on the base, we get
\[ \Rightarrow D = 4 - 4\left( 3 \right)\left( a \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow D = 4 - 12a\]
Now, we will find the discriminant of a quadratic equation \[{x^2} - 4x + 2 = 0\].
Comparing the equations \[{x^2} - 4x + 2 = 0\] and \[a{x^2} + bx + c = 0\], we get
\[a = 1\], \[b = - 4\], and \[c = 2\]
Substituting \[a = 1\], \[b = - 4\], and \[c = 2\] in the formula \[D = {b^2} - 4ac\], we get
\[ \Rightarrow D = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( 2 \right)\]
Applying the exponent on the base, we get
\[ \Rightarrow D = 16 - 4\left( 1 \right)\left( 2 \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow D = 16 - 8\]
Subtracting the terms in the expression, we get
\[ \Rightarrow D = 8\]
It is given that the discriminant of \[3{x^2} + 2x + a = 0\] is double the discriminant of \[{x^2} - 4x + 2 = 0\].
Therefore, we get the equation
\[ \Rightarrow 4 - 12a = 8\]
This is a linear equation in terms of \[a\]. We will solve this linear equation to find the value of \[a\].
Rewriting the expression, we get
\[ \Rightarrow 12a = 4 - 8\]
Subtracting the terms, we get
\[ \Rightarrow 12a = - 4\]
Dividing both sides of the equation by 12, we get
\[ \Rightarrow \dfrac{{12a}}{{12}} = \dfrac{{ - 4}}{{12}}\]
Thus, we get
\[ \Rightarrow a = - \dfrac{1}{3}\]
\[\therefore \] We get the value of \[a\] as \[ - \dfrac{1}{3}\].

Note: We used the term “quadratic equation” in our solution. A quadratic equation is an equation of degree 2. It is of the form \[a{x^2} + bx + c = 0\], where \[a\] is not equal to 0. A quadratic equation has 2 solutions.
We have formed a linear equation in one variable in terms of \[a\] in the solution. A linear equation in one variable is an equation that can be written in the form \[ax + b = 0\], where \[a\] is not equal to 0, and \[a\] and \[b\] are real numbers. For example, \[x - 100 = 0\] and \[100P - 566 = 0\] are linear equations in one variable \[x\] and \[P\] respectively. A linear equation in one variable has only one solution.