Question

Find a quadratic equation such that the roots of the equation are -7/2 and -3/5.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Quadratic\:eqn\to x^{2}+4.2x+2.1=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies \alpha = \frac{-7}{2} \\ \\ \tt: \implies \beta = \frac{-3}{5} \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies Quadratic \: eqn = ?$

• According to given question :

$\bold{For \: sum \: of \: zeroes} \\ \green{\tt: \implies \alpha + \beta = \frac{-7}{2} + \frac{-3}{5} } \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies \alpha \beta = \frac{-7}{2}\frac{-3}{5} \\ \\ \green{\tt: \implies \alpha \beta = 2.1 } \\ \\ \bold{For \: Quadratic \: eqn : } \\ \tt: \implies {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0 \\ \\ \tt: \implies {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ \\ {\tt: \implies {x}^{2} - ( \frac{-7}{2} + \frac{-3}{5})x + 2.1 = 0}$

$\tt: \implies {x}^{2} - (\frac{( - 7) \times 5 + ( - 3) \times 2}{10} )x + 2.1 = 0 \\ \\ \tt: \implies {x}^{2} + ( \frac{41}{10} )x + 2.1 = 0 \\ \\ \green{\tt: \implies {x}^{2} + 4.1x + 2.1 = 0}$

Answer 2

..

$\mathfrak{\huge{\purple{\boxed{\boxed{ Question \::- }}}}}$

Find a quadratic equation such that the roots of the equation are -7/2 and -3/5.

$\mathfrak{\huge{\green{\boxed{\boxed{ Solution \::- }}}}}$

In the above Question , the following information is given -

The roots of the Quadratic equation are -7/2 and -3/5.

Now,

A Quadratic Equation can be expressed as -

${ x } ^ 2 - sx + p = 0$

Where -

S refers to the sum of the roots .

P refers to the product of roots.

Here -

Sum Of Roots = ( - 7 / 2 ) - ( 3 / 5 )

=> Sum = ( 41 / 10 )

Product Of Roots = ( -7 / 2 ) × ( -3 / 5 )

=> Product = 2.1

Hence the required equation becomes -

${ x } ^ 2 - ( \dfrac{41}{10} ) x + \dfrac{ 21}{10} = 0 \\ \\ => 10 { x } ^ 2 - 41 x + 21 = 0$