Question

solve using quadratic formula

${abx}^{2} + ( {b}^{2} - ac)x - bc = 0$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm \: {abx}^{2} + ( {b}^{2} - ac)x - bc = 0$

We know,

The roots of the quadratic equation Ax² + Bx + C = 0, using quadratic formula is given by

$\boxed{\sf{  \:x = \frac{ - B \: \pm \: \sqrt{D} }{2A} \: }} \\ \\ \rm \: where \: D = {B}^{2} - 4AC$

Now, on comparing the given equation with Ax² + Bx + C = 0, we get

$\rm \: A = ab$

$\rm \: B = {b}^{2} - ac$

$\rm \: C = - bc$

Now, Consider Discriminant,

$\rm \: D = {B}^{2} - 4AC$

On substituting the values of A, B and C, we get

$\rm \: =  \: {( {b}^{2} - ac)}^{2} - 4(ab)( - bc)$

$\rm \: =  \: {b}^{4} + {a}^{2} {c}^{2} - {2b}^{2}ac + 4 {b}^{2}ac$

$\rm \: =  \: {b}^{4} + {a}^{2} {c}^{2} + {2b}^{2}ac$

$\rm \: = \: {( {b}^{2} + ac)}^{2}$

So,

$\rm\implies \:D\rm \: = \: {( {b}^{2} + ac)}^{2}$

Now, using quadratic formula, we have

$\rm \: x = \dfrac{ - B \: \pm \: \sqrt{D} }{2A}$

So, on substituting the values of B, A and D, we get

$\rm \: x = \dfrac{ - ( {b}^{2} - ac) \: \pm \: \sqrt{( {b}^{2} + ac)^{2} } }{2ab}$

$\rm \: x = \dfrac{ - ( {b}^{2} - ac) \: \pm \: ( {b}^{2} + ac)}{2ab}$

$\rm \: x = \dfrac{ - ( {b}^{2} - ac)+( {b}^{2} + ac)}{2ab} \: \: or \: \: x = \dfrac{ - ( {b}^{2} - ac) - ( {b}^{2} + ac)}{2ab}$

$\rm \: x = \dfrac{ - {b}^{2} + ac+{b}^{2} + ac}{2ab} \: \: or \: \: x = \dfrac{ - {b}^{2} + ac -{b}^{2} - ac}{2ab}$

$\rm \: x = \dfrac{ 2ac}{2ab} \: \: or \: \: x = \dfrac{ - 2{b}^{2}}{2ab}$

$\rm\implies \:\rm \: x = \dfrac{ c}{b} \: \: or \: \: x = \dfrac{ -{b}}{a}$

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

Step-by-step explanation:

$\color{olive}\[ \begin{array}{l} \text { Here, } \tt a b x^{2}+\left(b^{2}-a c\right) x-b c=0 \\ \\ \tt x=\dfrac{-B \pm \sqrt{\left(B^{2}-4 A C\right)}}{2 A} \\ \\ \tt \Rightarrow x=\dfrac{-\left(b^{2}-a c\right) \pm \sqrt{\left(b^{2}-a c\right)^{2}-4 a b(-b c)}}{2 a b} \\ \\ \tt\Rightarrow x=\dfrac{-\left(b^{2}-a c\right) \pm \sqrt{\left(b^{2}-a c\right)^{2}+4 a b^{2} c}}{2 a b} \\ \\ \tt\Rightarrow x=\dfrac{-\left(b^{2}-a c\right) \pm \sqrt{b^{4}-2 a b^{2} c+a^{2} c^{2}+4 a b^{2} c}}{2 a b} \\ \\ \tt \Rightarrow x=\dfrac{-\left(b^{2}-a c\right) \pm \sqrt{\left(b^{2}+a c\right)^{2}}}{2 a b} \\ \\ \tt \Rightarrow x=\dfrac{-\left(b^{2}-a c\right)+\left(b^{2}+a c\right)}{2 a b}, x=\dfrac{-\left(b^{2}-a c\right)-\left(b^{2}+a c\right)}{2 a b} \\ \\ \tt \Rightarrow x=\dfrac{2 a c}{2 a b} \quad, \quad \quad x = \dfrac{-2 b^{2}}{2 a b} \\ \\ \tt\Rightarrow x = \dfrac{c}{b} \qquad, \qquad x = - \dfrac{b}{a} \end{array} \]$