Question

Find the value of ‘ \[m\] ’ for which the given equation \[{x^2} + m\left( {2x + 1} \right) - 2x + 5 = 0\] has real and equal roots.

Answer 1

Hint: Quadratic equations can be expressed as a form \[a{x^2} + bx + c = 0\] , where \[a,b,c\] are constants. We solve the quadratic equation by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
First, we have to express the given quadratic equation in standard form \[a{x^2} + bx + c = 0\] , where \[a,b,c\] are constants. Then find the values of \[a,b,c\] from the given quadratic equation. Using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] find the roots say \[{x_1}\] and \[{x_2}\] . Given roots are equal i.e., \[{x_1} = {x_2}\] using this find the values of \[m\] .

Complete step-by-step answer:
Given \[{x^2} + m\left( {2x + 1} \right) - 2x + 5 = 0\] ---(1)
The equation (1) can be rewrite as
 \[{x^2} + \left( {2m - 2} \right)x + m + 5 = 0\] ---(2)
We solve the equation (2), using the formula
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] ---(3)
 In the equation (2) \[a = 1\] , \[b = 2\left( {m - 1} \right)\] and \[c = m + 5\] then the equation (3) becomes
 \[x = \dfrac{{ - 2\left( {m - 1} \right) \pm \sqrt {4{{\left( {m - 1} \right)}^2} - 4\left( 1 \right)\left( {m + 5} \right)} }}{{2\left( 1 \right)}}\]
Taking \[4\] common inside the square root, we get
 \[x = \dfrac{{ - 2\left( {m - 1} \right) \pm \sqrt {4\left\{ {{{\left( {m - 1} \right)}^2} - \left( {m + 5} \right)} \right\}} }}{2}\] ---(4)
Using the formula \[{\left( {a + b} \right)^2}\] the equation (4) becomes
 \[x = \dfrac{{ - 2\left( {m - 1} \right) \pm \sqrt {4\left\{ {{m^2} + 1 - 2m - m - 5} \right\}} }}{2}\] --(5)
Simplifying the equation (5), we get
 \[x = \dfrac{{ - 2\left( {m - 1} \right) \pm 2\sqrt {\left\{ {{m^2} + 1 - 2m - m - 5} \right\}} }}{2}\]
 \[ \Rightarrow \] \[x = \dfrac{{2\left\{ {1 - m \pm \sqrt {{m^2} - 3m - 4} } \right\}}}{2}\]
 \[ \Rightarrow \] \[x = 1 - m \pm \sqrt {{m^2} - 3m - 4} \] .
Let \[{x_1} = 1 - m + \sqrt {{m^2} - 3m - 4} \] and \[{x_2} = 1 - m - \sqrt {{m^2} - 3m - 4} \] be two roots of the given equation (1).
Given roots are real and equal
i.e., \[{x_1} = {x_2}\]
 \[ \Rightarrow \] \[1 - m + \sqrt {{m^2} - 3m - 4} = 1 - m - \sqrt {{m^2} - 3m - 4} \]
 \[ \Rightarrow \] \[ + \sqrt {{m^2} - 3m - 4} = - \sqrt {{m^2} - 3m - 4} \]
 \[ \Rightarrow \] \[2\sqrt {{m^2} - 3m - 4} = 0\]
Since \[2 \ne 0\]
 \[ \Rightarrow \] \[\sqrt {{m^2} - 3m - 4} = 0\]
Squaring on both sides, we get
 \[{m^2} - 3m - 4 = 0\] --(6)
The equation (6) is in quadratic form
 \[ \Rightarrow \] \[m = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 1 \right)\left( { - 4} \right)} }}{{2\left( 1 \right)}}\]
 \[ \Rightarrow \] \[m = \dfrac{{3 \pm \sqrt {9 + 16} }}{2}\]
 \[ \Rightarrow \] \[m = \dfrac{{3 \pm 5}}{2}\]
 \[ \Rightarrow \] \[m = \dfrac{{3 + 5}}{2},\dfrac{{3 - 5}}{2}\]
 \[ \Rightarrow \] \[m = 4, - 1\] .
Hence the values of \[m\] are \[4\] and \[ - 1\] for which the given equation \[{x^2} + m\left( {2x + 1} \right) - 2x + 5 = 0\] has real and equal roots.
When \[m = 4\] then the equation (1) can be written as
 \[{x^2} + 6x + 9 = 0\] and the repeated roots are \[ - 3, - 3\] .
When \[m = - 1\] then the equation (1) can be written as
 \[{x^2} - 4x + 4 = 0\] and the repeated roots are \[2,2\] .

Note: Note that the standard form, factored form and vertex form are three forms a quadratic form should be written. \[a{x^2} + bx + c = 0\] where \[a\] , \[b\] and \[c\] are constants is the equation in standard form. \[\left( {ax + c} \right)\left( {bx + d} \right) = 0\] where \[a\] , \[b\] , \[d\] and \[c\] are constants is the equation in standard form. \[a{\left( {x + b} \right)^2} + c = 0\] where \[a\] , \[b\] and \[c\] are constants, is the equation in vertex form. The quadratic equation is used to find the curve on a Cartesian grid. It is primarily used to find the curve that objects take when they fly through the air.