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

Last updated date: 18th Mar 2023

•

Total views: 306k

•

Views today: 7.85k

Answer

Verified

306k+ views

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India