Question

⇒ For what value ok K will the following equation give only one solution? Also find the solution for that value of K.
⇒ 3x² + Kx + 2 = 0
⇒ Solution : √2/√3
⇒ Solution for K : -√2/√3

Answer 1

➤ Result

Two equal roots are

• $x=\dfrac{\sqrt{6} }{3}$ if $K=-2\sqrt{6}$
• $x=-\dfrac{\sqrt{6} }{3}$ if $K=2\sqrt{6}$

➤ Definition of Discriminant

Let's attempt to find quadratic formula, via complete the square method.

☆ Complete the Square

Given: A quadratic equation $ax^2+bx+c=0$

$\implies x^2+\dfrac{b}{a} x+\dfrac{c}{a} =0$

$\implies x^2+\dfrac{b}{a} x+\dfrac{c}{a} +\dfrac{b^2}{4a^2} =\dfrac{b^2}{4a^2}$

$\implies x^2+\dfrac{b}{a} x+\dfrac{b^2}{4a^2} =\dfrac{b^2}{4a^2} -\dfrac{c}{a}$

$\implies (x+\dfrac{b}{2a} )^2=\dfrac{b^2-4ac}{4a^2}$

☆ Definition of the Square Root

The number which gives $x+\dfrac{b}{2a}$ after squaring is $\pm\sqrt{\dfrac{b^2-4ac}{4a^2}$.

$\therefore x+\dfrac{b}{2a} =\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$

$\therefore x+\dfrac{b}{2a} =\pm\dfrac{\sqrt{b^2-4ac}}{2a}$

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

Hence, discriminant $D=b^2-4ac$.

☆Nature of the Roots

Nature of the roots depends on the value of the square root.

Hence

• $D>0$ then real and different roots.
• $D=0$ then real and equal roots.
• $D<0$ then real and different roots.

→ Hence we arrive at the conclusion that $D=0$ is satisfied.

➤ Application

So, let's apply this to our question.

Given: $3x^2+Kx+2=0$

Via zero discriminant, we get

$K^2-24=0$

$\implies K=\pm 2\sqrt{6}$

Therefore we need to solve $3x^2\pm2\sqrt{6} x+2=0$ to get our solutions.

☆ Calculation

Let's focus on that this equation is derived from zero discriminant.

Hence, equal roots.

The roots can be found using

• Vieta's Formula. (Sum of two roots.)
• Arithmetic Mean. (Two equal roots.)

$\alpha +\beta =-\dfrac{\pm 2\sqrt{6} }{3}$ [Vieta's Formula]

$\implies \dfrac{\alpha +\beta }{2} =\mp \dfrac{\sqrt{6} }{3}$ [Arithmetic Mean]

Hence, two equal roots are

• $x=\dfrac{\sqrt{6} }{3}$ if $K=-2\sqrt{6}$
• $x=-\dfrac{\sqrt{6} }{3}$ if $K=2\sqrt{6}$

➤ More information

☆ Fundamental Theorem of Algebra

An n-th degree polynomial has exactly n roots. So, the number of roots in the quadratic equation is 2.

☆ Vieta's Formula

Vieta's formula can be proved by factor theorem.

If we consider $x^2+\dfrac{b}{a} x+\dfrac{c}{a} =0$, which roots are $x=\alpha ,\beta$

$x^2+\dfrac{b}{a} x+\dfrac{c}{a} =(x-\alpha )(x-\beta )$

Hence

$x^2+\dfrac{b}{a} x+\dfrac{c}{a} =x^2-(\alpha +\beta )x+\alpha \beta$

The results are:

• Sum of the roots $\alpha +\beta =-\dfrac{b}{a}$
• Product of the roots $\alpha \beta =\dfrac{c}{a}$

Answer 2

Step-by-step explanation:

GIVEN

We are given a quadratic equation which contains an unknown variable K.

==>3x^2+ Kx+2=0

TO FIND

We need to find a value of k for which this equation has only one root

We know that for a quadratic equation to have a single solution, The condition would be that discriminant should be equal to 0

==> b^2-4ac = 0

PROCEDURE

3x^2 +Kx +2 =0

==> k^2 - 4(3)(2) = 0

==> k^2- 24= 0

==> k^2 =24

==> k = ✓24

==> k= 2✓6 or -2✓6

CASE 1

When k = 2✓6

Given Q.E becomes 3x^2+2✓6x+2 =0

Using the Sridacharya formula , we get,

==>x = (-b +- ✓D)/2a

==> -2✓6 /6

==> -✓2/✓3 is the solution

CASE 2

When k = -2✓6

The Q.E becomes 3x^2-2✓6x+2=0

By following the same procedure as above , we get

==> x = √2/✓3 is the solution.

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