Question

Find the quadratic equation whose roots exceed the roots of the quadratic equation \[{x^2} - 2x + 3 = 0\] by\[2\].

Answer 1

Hint: Quadratic equations are the equation that contains at least one squared variable which is equal to zero. Quadratic equations are useful in our daily life; they are used to calculate areas, speed of the objects, projection, etc. Quadratic equation is given as \[a{x^2} + bx + c = 0\].This is the basic equation which contains a squared variable \[x\]and three constants a, b and c. The value of the \[x\] in the equation which makes the equation true is known as the roots of the equation. The numbers of roots in the quadratic equations are two as the highest power on the variable of the equation is x. The roots of the equation are given by the formula\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where \[{b^2} - 4ac\] tells the nature of the solution. In the quadratic equation, \[a{x^2} + bx + c = 0\] the sum of the roots is given by \[ - \dfrac{b}{a}\]whereas their products is given \[\dfrac{c}{a}\].

Complete step by step solution: In the given quadratic equation \[{x^2} - 2x + 3 = 0\]--- (i)
Let us assume that the roots of the equation to be \[\alpha \] and \[\beta \].
Hence as we know the sum of the roots of the equation is given as\[\alpha + \beta = - \dfrac{b}{a}\]hence in this question, the sum will be \[\alpha + \beta = - \dfrac{b}{a} = - \dfrac{{\left( { - 2} \right)}}{1} = 2\] and the product will be\[\alpha \beta = \dfrac{c}{a} = \dfrac{3}{1} = 3\].
For the new quadratic equation the roots are exceeded by 2 so, \[\alpha + 2\]and\[\beta + 2\] are the new roots of the quadratic equation.
The sum of the roots of the new equation will be:
\[ \alpha ' + \beta ' = \left( {\alpha + 2} \right) + \left( {\beta + 2} \right) \\
   = \alpha + \beta + 4 \\
   = 2 + 4 \\
   = 6 \\ \]
As \[\alpha ' + \beta ' = \dfrac{{ - b}}{a} = 6\]
And the products of the roots is:
\[ \alpha '\beta ' = \left( {\alpha + 2} \right)\left( {\beta + 2} \right) \\
   = \alpha \beta + 2\alpha + 2\beta + 4 \\
   = \alpha \beta + 2\left( {\alpha + \beta } \right) + 4 \\
   = 3 + \left( {2 \times 2} \right) + 4 \\
   = 11 \\ \]
So \[\alpha '\beta ' = \dfrac{c}{a} = 11\]
Therefore new quadratic equation will be:
\[ {x^2} - \dfrac{b}{a}x + \dfrac{c}{a} = 0 \\
  {x^2} - 6x + 11 = 0 \\ \]
Hence the equation is \[{x^2} - 6x + 11 = 0\]

Note: In the quadratic equation if \[{b^2} - 4ac > 0\]the equation will have two real roots. If it is equal \[{b^2} - 4ac = 0\]then the equation will have only one real root and when \[{b^2} - 4ac < 0\]then the root is in complex form.