# Find the zeros of the following quadratic polynomial \[4{{\text{x}}^2} - 7x - 3\] and verify the relationship between the zeros and the coefficients.

Answer

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Hint- Here we will be using the actual sum and product of zeros of quadratic polynomials that we can find by Sridharacharya formula and later on compare with the formula for sum and product of roots of any quadratic equation.

Complete step-by-step answer:

We have, a quadratic equation \[4{{\text{x}}^2} - 7x - 3\]

Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] , where a not equal to 0 , & a, b, c are real coefficients of the equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]

Being quadratic it has 2 roots.

X = $\dfrac{{\left( { - {\text{b + }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}{\text{ & }}\dfrac{{\left( { - {\text{b - }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}$ --(1)

On comparing the given equation \[4{{\text{x}}^2} - 7x - 3\] with the general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] we got values of coefficients a = 4, b = -7, c = -3

On putting the value of coefficients a, b, c in equation (1)

${\text{x = }}\dfrac{{\left( { - ( - 7){\text{ + }}\sqrt {{{( - 7)}^2} - 4 \times (4) \times ( - 3)} } \right)}}{{2 \times 4}}{\text{ & }}\dfrac{{\left( { - ( - 7){\text{ - }}\sqrt {{{( - 7)}^2} - 4 \times (4) \times ( - 3)} } \right)}}{{2 \times 4}}$

${\text{x = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ & }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8}$

We know that if discriminant D $ \geqslant {\text{0}}$ then it will give real and distinct roots.

Here D= ${\text{ = }}\sqrt {{{( - 7)}^2} - 4 \times 4 \times ( - 3)} {\text{ = }}\sqrt {97} {\text{ }} \geqslant {\text{0}}$ Therefore we got two distinct real roots ${{\text{x}}_1}{\text{ = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ \& }}{{\text{x}}_2}{\text{ = }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8}$

Sum of roots \[{{\text{x}}_1}{\text{ + }}{{\text{x}}_2}{\text{ = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ + }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8} = {\text{ }}\dfrac{{7 + \sqrt {97} + 7 - \sqrt {97} {\text{ }}}}{8}{\text{ }}\]

On solving

Sum of roots = $\left( {\dfrac{{14}}{8}} \right) = {\text{ }}\left( {\dfrac{7}{4}} \right)$--- (2)

Product of roots = \[{{\text{x}}_1}{\text{ }} \times {\text{ }}{{\text{x}}_2}{\text{ = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ }} \times {\text{ }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8} = {\text{ }}\dfrac{{(7 + \sqrt {97} {\text{ )}} \times (7 - \sqrt {97} {\text{ ) }}}}{8}{\text{ }}\]

Using identity $\left( {{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}} \right){\text{ = (a + b)(a - b)}}$

\[{{\text{x}}_1}{\text{ }} \times {\text{ }}{{\text{x}}_2}{\text{ = }}\dfrac{{(7 + \sqrt {97} {\text{ )}} \times (7 - \sqrt {97} {\text{ ) }}}}{{8 \times 8}}{\text{ = }}\dfrac{{(49 - 97)}}{{64}} = \dfrac{{ - 48}}{{{\text{ }}64}}\]

On reducing the fraction

Product of roots = \[{{\text{x}}_1}{\text{ }} \times {\text{ }}{{\text{x}}_2} = \dfrac{{ - 48}}{{{\text{ }}64}} = \dfrac{{ - 3}}{{{\text{ 4}}}}\] --- (3)

We know that , In general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]

sum of roots equal to $\left( {\dfrac{{ - {\text{b}}}}{{\text{a}}}} \right)$ and product of roots equal to $\left( {\dfrac{{\text{c}}}{{\text{a}}}} \right)$

According to above results

In the given quadratic equation \[4{{\text{x}}^2} - 7x - 3\],

values of coefficients a = 4, b = -7, c = -3

On putting the values of coefficients

sum of roots = $\left( {\dfrac{{ - ( - 7)}}{4}} \right) = \dfrac{7}{4}$ --- (4)

and product of roots equal to $\left( {\dfrac{{ - 3}}{4}} \right)$--- (5)

According to equation 2 and 4, It has been proved that sum of zeros equal to $\left( {\dfrac{{ - {\text{b}}}}{{\text{a}}}} \right)$ which is a coefficient relation. Similarly by equation 3 and 5, It has been proved that product of zeros equal to $\left( {\dfrac{{\text{c}}}{{\text{a}}}} \right)$

Note- In these types of problems, we have to compare the given quadratic equation with the general quadratic equation and find the coefficient so that we can apply the sum of roots and product of roots formula. In order to find zeroes either use the Sridharacharya formula or factorise by the middle term split method.

Complete step-by-step answer:

We have, a quadratic equation \[4{{\text{x}}^2} - 7x - 3\]

Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] , where a not equal to 0 , & a, b, c are real coefficients of the equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]

Being quadratic it has 2 roots.

X = $\dfrac{{\left( { - {\text{b + }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}{\text{ & }}\dfrac{{\left( { - {\text{b - }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}$ --(1)

On comparing the given equation \[4{{\text{x}}^2} - 7x - 3\] with the general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] we got values of coefficients a = 4, b = -7, c = -3

On putting the value of coefficients a, b, c in equation (1)

${\text{x = }}\dfrac{{\left( { - ( - 7){\text{ + }}\sqrt {{{( - 7)}^2} - 4 \times (4) \times ( - 3)} } \right)}}{{2 \times 4}}{\text{ & }}\dfrac{{\left( { - ( - 7){\text{ - }}\sqrt {{{( - 7)}^2} - 4 \times (4) \times ( - 3)} } \right)}}{{2 \times 4}}$

${\text{x = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ & }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8}$

We know that if discriminant D $ \geqslant {\text{0}}$ then it will give real and distinct roots.

Here D= ${\text{ = }}\sqrt {{{( - 7)}^2} - 4 \times 4 \times ( - 3)} {\text{ = }}\sqrt {97} {\text{ }} \geqslant {\text{0}}$ Therefore we got two distinct real roots ${{\text{x}}_1}{\text{ = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ \& }}{{\text{x}}_2}{\text{ = }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8}$

Sum of roots \[{{\text{x}}_1}{\text{ + }}{{\text{x}}_2}{\text{ = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ + }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8} = {\text{ }}\dfrac{{7 + \sqrt {97} + 7 - \sqrt {97} {\text{ }}}}{8}{\text{ }}\]

On solving

Sum of roots = $\left( {\dfrac{{14}}{8}} \right) = {\text{ }}\left( {\dfrac{7}{4}} \right)$--- (2)

Product of roots = \[{{\text{x}}_1}{\text{ }} \times {\text{ }}{{\text{x}}_2}{\text{ = }}\dfrac{{\left( {{\text{7 + }}\sqrt {97} } \right)}}{8}{\text{ }} \times {\text{ }}\dfrac{{\left( {{\text{7 - }}\sqrt {97} } \right)}}{8} = {\text{ }}\dfrac{{(7 + \sqrt {97} {\text{ )}} \times (7 - \sqrt {97} {\text{ ) }}}}{8}{\text{ }}\]

Using identity $\left( {{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}} \right){\text{ = (a + b)(a - b)}}$

\[{{\text{x}}_1}{\text{ }} \times {\text{ }}{{\text{x}}_2}{\text{ = }}\dfrac{{(7 + \sqrt {97} {\text{ )}} \times (7 - \sqrt {97} {\text{ ) }}}}{{8 \times 8}}{\text{ = }}\dfrac{{(49 - 97)}}{{64}} = \dfrac{{ - 48}}{{{\text{ }}64}}\]

On reducing the fraction

Product of roots = \[{{\text{x}}_1}{\text{ }} \times {\text{ }}{{\text{x}}_2} = \dfrac{{ - 48}}{{{\text{ }}64}} = \dfrac{{ - 3}}{{{\text{ 4}}}}\] --- (3)

We know that , In general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]

sum of roots equal to $\left( {\dfrac{{ - {\text{b}}}}{{\text{a}}}} \right)$ and product of roots equal to $\left( {\dfrac{{\text{c}}}{{\text{a}}}} \right)$

According to above results

In the given quadratic equation \[4{{\text{x}}^2} - 7x - 3\],

values of coefficients a = 4, b = -7, c = -3

On putting the values of coefficients

sum of roots = $\left( {\dfrac{{ - ( - 7)}}{4}} \right) = \dfrac{7}{4}$ --- (4)

and product of roots equal to $\left( {\dfrac{{ - 3}}{4}} \right)$--- (5)

According to equation 2 and 4, It has been proved that sum of zeros equal to $\left( {\dfrac{{ - {\text{b}}}}{{\text{a}}}} \right)$ which is a coefficient relation. Similarly by equation 3 and 5, It has been proved that product of zeros equal to $\left( {\dfrac{{\text{c}}}{{\text{a}}}} \right)$

Note- In these types of problems, we have to compare the given quadratic equation with the general quadratic equation and find the coefficient so that we can apply the sum of roots and product of roots formula. In order to find zeroes either use the Sridharacharya formula or factorise by the middle term split method.

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