Question

Q3) The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.​

Answer 1

☆Given Question :-

• The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.

$\huge \orange{AηsωeR} ✍$

☆Given :-

• Sum of two roots of a quadratic equation is 5
• Sum of their cubes is 35

☆To Find :-

• The required Quadratic equation.

☆ Formula used :-

• If a and b are the roots of the quadratic equation, then the required quadratic equation is

$\sf \: ⟼ {x}^{2} - (a + b)x + ab = 0$

Solution :-

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{first \: root \: of \: quadratic \: equation \: be \: a} \\ &\sf{second \: root \: of \: quadratic \: equation \: be \: b} \end{cases}\end{gathered}\end{gathered}$

$\large \red{\bf \: According \: to \: statement } ✍$

☆ Sum of the roots of quadratic equation is 5

$\sf \: ⟼a + b = 5 \: ⟼ \: (1)$

$\large \red{\bf \: According \: to \: statement \: again} ✍$

☆ Sum of the cubes of the roots is 35.

$\sf \: ⟼ {a}^{3} + {b}^{3} = 35$

☆ Using identity, we get

$\sf \: ⟼ {(a + b)}^{3} - 3ab(a + b) = 35$

☆ On substituting the value of a + b, from equation (1), we get

$\sf \: ⟼ {(5)}^{3} - 3ab \times 5 = 35$

$\sf \: ⟼125 - 15ab = 35$

$\sf \: ⟼ - 15ab = 35 - 125$

$\sf \: ⟼ - 15ab = - 90$

$\sf \: ⟼ab = 6 \: ⟼ \: (2)$

Now,

☆ The required Quadratic equation having sum of roots (a + b) is 5 and product of roots (ab) is 6 is given by

$\sf \: ⟼ {x}^{2} - 5x + 6 = 0$