Question

Find the roots of equation, if they exists, by applying the quadratic formula :

$x {}^{2} + 5x - (a {}^{2} + a - 6) = 0$


Ans: x = ( a - 2 ) or x = - ( a + 3 )

Answer 1

Given:

x² + 5 x - ( a² + a - 6 ) = 0

==> x² + ( 2 + 3 ) x - ( a² + 3 a - 2 a - 6) = 0

==> x² + ( a + 2 + 3 - a ) x - ( a { a + 3 } - 2 { a + 3 } ) = 0

==> x² + ( a + 3 - a + 2 ) x - ( a - 2 ) ( a + 3 ) = 0

==> x² + ( a + 3 ) x - ( a - 2 ) x - ( a - 2 ) (a + 3 ) = 0

==> x [ x + a + 3 ] - ( a - 2 ) [ x + a + 3 ] = 0

==> [ x + a + 3 ] [ x - a + 2 ]  = 0

==>  Either x + a+ 3 =0

==> x = - a - 3

==> x = - ( a + 3 )

or ,

x - a + 2 = 0

==> x = a - 2

The values of x are  ( a - 2 ) or  - (a + 3 )

Comparing x² + 5 x - a² - a + 6 with a x ² + bx + x

we get:

a =  1

b = 5

c = -a² - a + 6

b² - 4 ac = 5 ² -  4 ( - a² - a + 6 )

==> 25 + 4 a² + 4 a - 24

==> 4 a² + 4 a + 1

==> ( 2a )² + 2 * 2a * 1 +(1)²

==> ( 2 a  + 1 )² ......................(1)

By quadratic formula we know that:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

From 1 :

$x = \frac{-5 \pm\sqrt{(2a+ 1 )^2 }}{2}$

$\implies x = \frac { - 5 \pm ( 2 a + 1 ) }{ 2 }$

When it is + :

$\implies x = \frac { - 5 + 1 + 2 a } { 2 }$

$\implies x = \frac{ 2 a - 4 }{2}$

$\implies x = a - 2$

or :

$\implies x = \frac{ - 5 - 2 a - 1 }{2}$

$\implies x=\frac{ - 2a - 6}{2}$

$\implies x= - a - 3$

The values of x are  ( a - 2 ) or  - (a + 3 ) by using quadratic formula method.

Hope it helps you

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Answer 2

Given Equation is x^2 + 5x - (a^2 + a - 6) = 0

Here, a = 1, b = 5, c = -a^2 - a + 6

The solutions are:

(i)

$=> x=\frac{-b+ \sqrt{b^2 - 4ac}}{2a}$

$=>\frac{-5+\sqrt{(5)^2 - 4(1)(-a^2 - a + 6)}}{2}$

$=>\frac{-5+\sqrt{25+4a^2+4a-24}}{2}$

$=>\frac{-5+ \sqrt{4a^2 + 4a+1}}{2}$

$=>\frac{-5+ \sqrt{(2a + 1)^2}}{2}$

$=>\frac{-5+2a + 1}{2}$

$=>\frac{2a-4}{2}$

$=>\frac{2(a-2)}{2}$

$=> a-2$

(ii)

$=>x =\frac{-b- \sqrt{b^2-4ac}}{2a}$

$=>\frac{-5-\sqrt{(5)^2 - 4(1)(-a^2-a+6)}}{2a}$

$=>\frac{-5-\sqrt{25+4a^2 + 4a-24}}{2}$

$=>\frac{-5-\sqrt{(2a + 1)^2}}{2}$

$=>\frac{-5-2a-1}{2}$

$=>\frac{-2a-6}{2}$

$=>-\frac{2(a+3)}{2}$

$=>-(a+3)$

The roots of equation are:

$=>x=\boxed{(a - 2), -(a + 3)}$

Hope it helps!