Question

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

Answer 1

Step-by-step explanation:

$let \: a \:and \: b \\ then \\ a + b = 5 \\ \ \\ {a}^{3} + {b}^{3} = 35 \\ \\ by \: solving \: a = 2 \\and \: \: \: \: \: \: \: \: \: \: \: \: \: b = 3 \\ than \: equation \: \\ \\ (x - 2)(x - 3) = 0 \\ \\ {x}^{2} - 5x + 6 = 0$

Answer 2

Answer:

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

Step-by-step explanation:

Let the roots of Quadratic equation are

$\alpha \: and \: \beta$

According to the question

$\alpha + \beta = 5...eq1 \\ \\ { \alpha }^{3} + { \beta }^{3} = 35...eq2$

To find the value of roots take cube of eq1

${( \alpha + \beta )}^{3} = 125 \\ \\ open \: identity \\ \\ { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta ) = 125 \\ \\ 35 + 3 \alpha \beta (5) = 125 \\ \\ 15 \alpha \beta = 125 - 35 \\ \\ \alpha \beta = \frac{ 90}{15} \\ \\ \alpha \beta =6$

Now we have sum of roots as well as product of roots.

According to the relation of coefficient of polynomial and roots of Quadratic equation

$\alpha + \beta = \frac{ - b}{a} = 5 \\ \\ \alpha \beta = \frac{c}{a} =6$

polynomial is

${x}^{2} - \bigg(\frac{ - b}{a}\bigg) x + \frac{c}{a} = 0 \\ \\ {x}^{2} - (5)x + 6 = 0 \\ \\ {x}^{2} - 5x + 6 = 0$

Hope it helps you.