Question

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

give it correct ...fastt..please..​

Answer 1

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

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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Answer 2

Step-by-step explanation:

hope it's correct and helps you... : )