Math, asked by LavanyaThakur1234, 2 months ago

kx2 - 10x+3=0 is 1/3 root is given so plz solve and send​

Answers

Answered by Anonymous
39

\sf{Answer}

Given 1/3 is a root of Q.E kx² - 10x + 3 = 0

Hence it is aroot If we Substitute It should be equal to RHS

k (1/3)² - 10(1/3) + 3 = 0

k (1/9) - 10/3 + 3 = 0

k/9 - 10/3 + 3 =0

Take LCM to denominator

k - 30 + 27 /9 = 0

Do cross multiplication

k - 30 + 27 = 0

k - 3 = 0

k = 3

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Know more

General form of Quadratic equation is

ax² + bx + c = 0

A quadratic equation has 2 roots

We can find those two roots by different methods like

FACTORISATION METHOD

QUADRATIC FORMULA

COMPLETE SQUARING etc

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Nature of roots :-

It can be determined by discriminant of Quadratic equation

Nature of roots :- Whether the roots are complex , equal , real

It can be determined by discriminant

D = b² - 4ac

D > 0 roots are real

D< 0 roots are complex & conjugate

D = 0 roots are real &equal

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Answered by llMrIncrediblell
410

\underline{\underline{\sf{\maltese\:\:Given}}}

  • kx² - 10x + 3 where x =  \frac{1}{3}

\underline{\underline{\sf{\maltese\:\:To\: Find}}}

  • The value of k in the given quadratic equation.

\underline{\underline{\sf{\maltese\:Calculations \:}}}

So, x =  \frac{1}{3} must satify the given equation.

 \longrightarrow \: k {x}^{2}  - 10x + 3 = 0

Now, putting x =  \frac{1}{3} in the given equation, we get :-

 \longrightarrow \: k( \frac{1}{3} ) ^{2} - 10 (\frac{1}{3}  ) + 3 = 0

 \longrightarrow \: k( \frac{1}{9} )  -  \frac{10}{3}  + 3 = 0

  \longrightarrow \: \frac{k}{9}  -  \frac{10}{3}  + 3 = 0

  \longrightarrow \: \frac{k   - 30 + 27}{9}  = 0

  \longrightarrow \: \frac{k   - 3}{9}  = 0

cross multiplying 9 by 0 to get :-

  \longrightarrow \: k   - 3 = 0 \times 9

 \longrightarrow \: k - 3 = 0

 \longrightarrow \: k = 3

Hence, the value of k is 3.

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\tt\pink{\underline{VERIFICATION}} :

Lets verify it by putting the value of k = 3 and x =  \frac{1}{3} , our answer is correct when the value of LHS is equal to RHS .

L.H.S

 \longrightarrow \: k {x}^{2}  - 10x + 3 = 0

substituting the values,

\longrightarrow \:3( \frac{1}{3} )^{2}  - 10( \frac{1}{3} ) + 3 = 0

\longrightarrow \:3( \frac{1}{9} ) -  \frac{10}{3}  + 3

\longrightarrow \: \cancel{3}( \frac{1}{ \cancel{9}} ) -  \frac{10}{3}  + 3

\longrightarrow \: \frac{1}{3}  - \frac{10}{3}  + 3

\longrightarrow \: \frac{1 - 10 + 9}{3}

\longrightarrow \: \frac{0}{3}

\longrightarrow \:0

LHS = RHS

Hence, the solution is verified.

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