Question

Solve the following quadratic equation:
${x}^{2} + x + \frac{1}{ \sqrt{2} } = 0$​

Answer 1

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

Given quadratic equation is

$\rm :\longmapsto\: {x}^{2} + x + \dfrac{1}{ \sqrt{2} } = 0$

can be rewritten as

$\rm :\longmapsto\: \sqrt{2} {x}^{2} + \sqrt{2}x + 1 = 0$

On comparing with ax² + bx + c = 0, we get

$\red{\rm :\longmapsto\:a = \sqrt{2}}$

$\red{\rm :\longmapsto\:b = \sqrt{2}}$

$\red{\rm :\longmapsto\:c = 1}$

So, Let we first evaluate the Discriminant, D which is given by D = b² - 4ac

So, on substituting the values, we get

$\rm :\longmapsto\:D = {( \sqrt{2} )}^{2} - 4 \times \sqrt{2} \times 1$

$\rm :\longmapsto\:D = 2 - 4\sqrt{2}$

Since, D < 0, it means equation have unreal or imaginary roots, which is evaluated as

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{D} }{2a}$

$\rm :\longmapsto\:x = \dfrac{ - \sqrt{2} \: \pm \: \sqrt{2 - 4 \sqrt{2} } }{2 \sqrt{2} }$

$\rm :\longmapsto\:x = \dfrac{ - \sqrt{2} \: \pm \: \sqrt{ (\sqrt{2}) (\sqrt{2}) - 4 \sqrt{2} } }{2 \sqrt{2} }$

$\rm :\longmapsto\:x = \dfrac{ - \sqrt{2} \: \pm \: i\sqrt{2} \sqrt{2 \sqrt{2} - 1} }{2 \sqrt{2} }$

$\rm :\longmapsto\:x = \dfrac{ - 1\: \pm \: i \sqrt{2 \sqrt{2} - 1} }{2}$

$\bf\implies \:\boxed{\tt{ \:x = \dfrac{ - 1\: \pm \: i \sqrt{2 \sqrt{2} - 1} }{2} \: }}$

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Explore More :-

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

Answer:

Hope the picture helps you ammu akka.