Question

Find the zeroes of the quadratic polynomial and verify
the relation ship between the 3x²-10x+3 zeroes and the
coefficient

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

here is the answer

Step-by-step explanation:

Let the given Polynomial be denoted by f(x) . Than ,

\begin{gathered}f(x) = \sqrt{3} {x}^{2} + 10x + 7 \sqrt{3} \\ = > \sqrt{3} {x}^{2} + 3x + 7x + 7 \sqrt{3} \\ = > \sqrt{3}x(x + \sqrt{3} ) + 7(x + \sqrt{3} ) \\ = > (x + \sqrt{3} )( \sqrt{3} x + 7) \\ \\ f(x) = 0 \\ \\ = > (x + \sqrt{3} ) = 0 \: \: or \: \: ( \sqrt{3} x + 7) = 0 \\ = > x = - \sqrt{3} \: \: \: \: \: \: \: \: and \: \: x = \frac{ - 7}{ \sqrt{3} }\end{gathered}f(x)=3x2+10x+73=>3x2+3x+7x+73=>3x(x+3)+7(x+3)=>(x+3)(3x+7)f(x)=0=>(x+3)=0or(3x+7)=0=>x=−3andx=3−7

so the zeros of f (x) are -√3 and -7/√3

We have to verify

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\begin{gathered}sum \: \: of \: zeros \: = ( - \sqrt{3} + \frac{ - 7}{ \sqrt{3} } ) = \frac{ - 10}{ \sqrt{3} } = \frac{coefficient \: \: of \: x)}{coefficiet \: \: of \: {x}^{2} } \\ \\ product \: \: of \: zeros = \frac{ - 7}{ \sqrt{3} } \times - \sqrt{3} = 7 = \frac{constant \: \: term}{coefficient \: \: of \: {x}^{2} }\end{gathered}sumofzeros=(−3+3−7)=3−10=coefficietofx2coefficientofx)productofzeros=3−7×−3=7=coefficientofx2constantterm

Hence verified ..

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