Question

If a and b are the zeros of the given quadratic polynomial f(x)= 5x2 - 7x + 1, find the
value 1/a + 1/b

Answer 1

Given that a and b are the zeroes of the polynomial f(x) = 5x² - 7x + 1

here, we've to find the value of

1/a + 1/b

so first of all, we needa find a and b (zeroes of the polynomial)

standard form of quadratic equation = ax² + bc + c = 0

here, a = 5, b = -7 and c = 1

by using quadratic formula, we get

$\tt x = \frac{ - b± \sqrt{( {b}^{2} - 4ac) } }{2a} \\ \\ \tt = \frac{ - ( - 7)± \sqrt{( { - 7})^{2} -( 4 \times 5 \times 1)} }{(2 \times 5)} \\ \\ \tt = \frac{7± \sqrt{49 - 20} }{10} \\ \\ \tt = \frac{7± \sqrt{29} }{10}$

therefore a = (7 + √29)/10 and b = (7 - √29)/10

hence, the value of 1/a + 1/b :-

= 1/(7 + √29) + 1/(7 - √29)

$\tt \scriptsize= \frac{10(7 - \sqrt{29} )}{(7 + \sqrt{29} )(7 - \sqrt{29} )} + \frac{10(7 + \sqrt{29} )}{(7 + \sqrt{29})(7 - \sqrt{29} )} \\ \\ \tt = \frac{70 - 10\sqrt{29} }{( {7})^{2} - ( { \sqrt{29} })^{2} } + \frac{70 + 10\sqrt{29} }{( {7})^{2} - ( { \sqrt{29} })^{2} } \\ \\ \tt = \frac{70 - \cancel{10\sqrt{29} } + 70 + \cancel{10\sqrt{29}} }{49 - 29} \\ \\ \tt = \frac{140}{20} = \boxed{ 7 }$

Answer 2

Answer :

$\textsf{The value of }\dfrac{1}{a}+\dfrac{1}{b}\textsf{is 7}$

Step-by-step explanation :

Given that :

If a and b are the zeros of the quadratic polynomial  :

$f(x)=5x^2-7x+1$

To Find :

$\textsf{The value of }\dfrac{1}{a}+\dfrac{1}{b}$

Solution :

$f(x)=5x^2-7x+1$

We know that :

$\bigstar\;\textsf{\underline{\underline{Sum of zeroes - }}}}\\\\\it \implies a+b=\dfrac{-b}{a}=\dfrac{7}{5}\\\\\bigstar \textsf{\underline{\underline{Product of zeroes - }}}\\\\\it \implies abc=\dfrac{c}{a}=\dfrac{1}{5}$

Put the given values :

$\tt \implies \dfrac{1}{a}+\dfrac{1}{b}\\\\\implies \dfrac{b+a}{ab}\\\\\implies \dfrac{\dfrac{7}{5}}{\dfrac{1}{5}}=7$

Hence,

The Correct answer is 7.