Question

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Solve -

i) 3x^2 - 2x + 1/3 = 0

Find the zeroes and verify the relationship with coeffcients.

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Answer 1

Answer:

So, here we are given the question, we have to find the zeroes and verify the relationship with coeffcients. For that,

Before solving this, lemme tell you that we can solve this by using two methods.

• >> By Using Quadratic Formula.

• >> By Using Method of Splitting Middle Term.

So, In the question as you mentioned the chapter name Quadratic Equation. Then we have to solve it using it.

Solve :

$:\implies \sf{p(x) = 3x^{2} - 2x + \dfrac{1}{3}}$

Solution :

Step by Step :

By Taking L.C.M.

$:\implies \sf{p(x) = 3x^{2} - 2x + \dfrac{1}{3}}$

$:\implies \sf{p(x) = \dfrac{3x^{2}}{1} - \dfrac{2x}{1} + \dfrac{1}{3} = 0}$

$:\implies \sf{p(x) = \dfrac{9x^{2} - 6x +1}{3} = 0}$

Transposing ' 3 ' on the R.H.S

Here, it is in division with p(x) = 9x² - 6x + 1 , by transposing it get into the multiplication with 0 which is on the right hand side.

Then,

$:\implies \sf{p(x) = 9x^{2} - 6x +1 = 0 \times 3}$

$:\implies \sf{p(x) = 9x^{2} - 6x +1 = 0 }$

Now,

Let us we consider the quadratic equation be -

$:\implies\sf ax^{2} + bx + c = 0$ { Where , a ≠ 0 }

$:\implies \sf{p(x) = 9x^{2} - 6x +1 = 0 }$

Then ,

Discriminant :

$:\implies\sf \beta = \dfrac{- b - \sqrt D}{2a}$

The value of the D is = $\sf b^{2} - 4 ac$ ≥ 0.

• a = 9

• b = - 6

• c = 1

Substituting the values, we get :

→ D = (-6) ² - 4 × 9 × 1

→ D = 36 - 4 × 9 × 1

→ D = 36 - 36

→ D = 0

Our Discriminant comes out to be zero, that means both the zeroes are equal.

$\bullet \: \:$ If D = $\sf b^{2} - 4 ac$ = 0 i.e, the discriminant of equation is zero, then equation has real and equal roots both equal to $\sf - \dfrac{b^{2}}{2a }$

Now, By using quadratic Formula :

$:\implies\sf \alpha = \dfrac{- b + \sqrt D}{2a}$

$:\implies\sf \alpha = \dfrac{- (-6) + 0}{2(9)}$

$:\implies\sf \alpha = \dfrac{6}{18}$

$:\implies\sf \alpha = \dfrac{\not{6}}{\not{18}}$

$:\implies\sf \alpha = \dfrac{1}{3}$

And from here, we got the value of one zeroes and we know both equal.

________________

Alternate Method

Use Split Middle term Method :

• Multiple the coefficient of x² with the constant.

• According to the Split Middle term Method - The middle term of the given expression will need to be split in such a way that their sum is equal to the middle term and their product is equal to the product of coefficient of x² and constant term.

• Now, taking common between two terms and making pair.

→ 9x² - 6x + 1 = 0

→ 9x² - (3 + 3)x + 1 = 0

Opening brackets :

9x^{2} - 3x - 3x + 1 = 0

Taking common :

→ 3x(3x - 1) - 1(3x - 1) = 0 }

→ (3x - 1) (3x - 1) = 0

Putting each term equal to zero.

→ 3x - 1 = 0

Transposing 1 and 3

$:\implies \sf{x = \dfrac{1}{3}}$

As we notice both terms are same. Then, both the zeroes are equal i.e, $\sf \dfrac{1}{3}$

Then, Considering one of them as $\sf \alpha$ and the other one as $\sf \beta$

$:\implies {\sf{ \alpha = {\purple{\dfrac{1}{3}}} \: and \: \beta \: also = {\purple{\dfrac{1}{3}}}}}$

Then,

$:\implies \sf{p(x) = 3x^{2} - 2x + \dfrac{1}{3}}$

Now,

Let's compare it with the general form of quadratic equation :

$\bold{Quadratic \: Equation: \: ax^2+bx+c=0}$

• a = 3
• b = -2
• $\sf { c =\dfrac{1}{3}}$

Verifying relation between zeroes and coefficients

$\dashrightarrow \: \: \sf{Sum \: of \: the \: Zeroes = - \dfrac{b}{a}}$

Where,

• b = Coefficient of x

• a = Coefficient of x²

So,

$\longrightarrow \sf (Sum \: of \: the \: Zeroes) = - \dfrac{Coefficient \: of \: x}{Coefficient \: of \: x^{2}}$

That is ,

$\implies \sf {\alpha + \beta} = {\red{\dfrac{Coefficient \: of \: x}{Coefficient \: of \: x^{2}}}}$

Plugging in the respective values, we get :

$\implies \sf \dfrac{1}{3} + \dfrac{1}{3} = - \dfrac{(-2)}{3}$

$\implies \sf \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}$

Taking L.C.M on the L.H.S side,

$\implies \sf \dfrac{1 + 1}{3} = \dfrac{2}{3}$

$\implies{\sf {\blue{\dfrac{2}{3} = \dfrac{2}{3}}}}$

${\underline{\bf{L.H.S = R.H.S}}}$

Hence,Verified !!

Now,

$\dashrightarrow \: \: \sf{Product \: of \: the \: Zeroes = \dfrac{c}{a}}$

Where,

• c = Constant term

• a = Coefficient of x²

$\dashrightarrow \: \: \sf{Product \: of \: the \: Zeroes = \dfrac{Constant \: Term}{Coefficient \: of \: x^{2}}}$

$\dashrightarrow \: \: \sf{\alpha \times \beta = {\red{\dfrac{Constant \: Term}{Coefficient \: of \: x^{2}} }}}$

$:\implies \sf{\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{\dfrac{1}{3}}{3}}$

$:\implies \sf{\dfrac{1}{9} = \dfrac{1}{3} \times 3 }$

$:\implies \sf{\blue{\dfrac{1}{9} = \dfrac{1}{9}}}$

${\underline{\bf{L.H.S = R.H.S}}}$

Hence, Verified !!

___________________

Answer 2

Given:

$3 {x}^{2} - 2x + 1 \div 3 = 0 \\ 9 {x}^{2} - 6x + 1 = 0 \\ {(3x - 1)}^{2} = 0 \\ x = \frac{1}{3} \: or \: \frac{1}{ 3}$

now

$x + y = 1 \div 3 + 1 \div 3 \\ = 2 \div 3 = - b \div a$

$xy = 1 \div 9 \\ = \frac{c}{a}$

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