Question

Solve the following quadratic equation using completing the square. Leave your answer in surd form.

1) 15 – 6x – 2x² = 0

Answer should be:–
x=
$\frac{ - 3 + \sqrt{39} }{2}$
AND
x=
$\frac{ - 3 - \sqrt{39} }{2}$
​

Answer 1

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

Given quadratic equation is

$\rm :\longmapsto\:15 - 6x - {2x}^{2} = 0$

can be rewritten as

$\rm :\longmapsto\: {2x}^{2} + 6x - 15 = 0$

$\rm :\longmapsto\: {2x}^{2} + 6x = 15$

Step :- 1 Make the coefficient of x² unity, divide the whole equation by 2

$\rm :\longmapsto\: {x}^{2} + 3x = \dfrac{15}{2}$

Step :- 2 Add the square of half the coefficient of x on both sides, we get

$\rm :\longmapsto\: {x}^{2} + 3x + {\bigg[\dfrac{3}{2} \bigg]}^{2} = \dfrac{15}{2} + {\bigg[\dfrac{3}{2} \bigg]}^{2}$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} + 2 \times \bigg[\dfrac{3}{2} \bigg] \times x + {\bigg[\dfrac{3}{2} \bigg]}^{2} = \dfrac{15}{2} + {\bigg[\dfrac{3}{2} \bigg]}^{2}$

We know,

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}$

So, using this

$\rm :\longmapsto\: {\bigg[x + \dfrac{3}{2} \bigg]}^{2} = \dfrac{15}{2} + \dfrac{9}{4}$

$\rm :\longmapsto\: {\bigg[x + \dfrac{3}{2} \bigg]}^{2} = \dfrac{30 + 9}{4}$

$\rm :\longmapsto\: {\bigg[x + \dfrac{3}{2} \bigg]}^{2} = \dfrac{39}{4}$

$\rm :\longmapsto\:x + \dfrac{3}{2} = \: \pm \: \dfrac{ \sqrt{39} }{2}$

$\rm :\longmapsto\:x = - \dfrac{3}{2} \pm \: \dfrac{ \sqrt{39} }{2}$

$\bf \implies\:x = \dfrac{ - 3 + \sqrt{39} }{2} \: \: \: or \: \: \: x = \dfrac{ - 3 - \sqrt{39} }{2}$

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