Question

Form a quadratic equation whose roots are -1/3 and 5/2​

Answer 1

Heya,

According to the question,

Roots of quadratic equation are 2 + √5 and 2 - √5

Let α = 2 + √5

ß = 2 - √5

Sum of zeros of polynomial = α + ß

= 2 + √5 + 2 - √5

= 2 + 2

= 4

Product of zeros of polynomial = αß

= (2 + √5)(2 - √5)

= 4 - 5

= -1

Quadratic polynomial = k(x² - (α + ß)x + αß)

=> k(x² - 4x - 1)

'k' is constant.

Hope this helps....:)

Answer 2

Quadratic equation of the given roots will be ${6x^{2} -13x-5}=0$

Step-by-step explanation:

Given roots are : $-\frac{1}{3}\ and \ \frac{5}{2}$

$= > Let\ a = \frac{-1}{3} \\\\b= \frac{5}{2}$

=> Adding a and b, we get :

$a+b= -\frac{1}{3} + \frac{5}{2}\\\\a+b = \frac{-2+15}{6}\\\\ a+b= \frac{-13}{6}$

=> Multiplying a and b, we get :

$a.b= -\frac{1}{3} * \frac{5}{2}\\\\a.b = \frac{-5}{6}$

=> We know that, a quadratic equation can be written as :

$x^{2} - (a+b)x+(a.b) =0$

Substituting the value of a+b and a.b in the above equation,

$x^{2} +( \frac{-13}{6})x + (\frac{-5}{6} )=0$

Taking LCM as 6 , we get :

$\frac{6x^{2} -13x-5}{6}=0$

${6x^{2} -13x-5}=0$

∴ The quadratic equation will be ${6x^{2} -13x-5}=0$