Question

The quadratic equation $x^2+3x+50 = 0$ has roots r and s. Find a quadratic question whose roots are r^2 and s^2. no spam and plz answer correct+ no copy
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Answer 1

Step-by-step explanation:

Given,

the quadratic equation is :

$\sf \: {x}^{2} + 3x + 50 = 0$

r and s are the roots of that equation.

so,

$\sf \:sum \: of \: roots \: \: \pink{r + s = - \frac{3}{1} = - 3 } \: \: \: \: and \\ \sf \: products \: of \: roots \: \: \: \: \: \: \: \: \pink{rs = \frac{50}{1} = 50}$

we know that,

General formula of quadratic equation is :

$\small{ \green{ \bf {x}^{2} - (sum \: of \: roots)x + poducts \: of \: roots = 0}}$

$\sf \: {r}^{2} \: \: and \: {s}^{2} \: are \: \: roots \: of \: a \: new \: equation \:$

Now, sum of roots of new equation is :

$\sf \: {r}^{2} + {s}^{2} = {(r + s)}^{2} - 2rs \\ \sf = {( - 3)}^{2} - 2 \times 50 \\ = 9 - 100 \\ = - 91$

According to the question, our quadratic equation is :

$\bf {x}^{2} - ( {r}^{2} + {s}^{2} )x + {r}^{2} {s}^{2} = 0 \\ \bf \implies \: {x}^{2} - ( - 91)x + {(rs)}^{2} = 0 \\ \bf \implies \: {x}^{2} + 91x + {(50)}^{2} = 0 \\ \bf \implies \: {x}^{2} + 91x + 2500 = 0$

Answer 2

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