How do you find a quadratic polynomial with integer coefficients which has #x=3/5+-sqrt29/5# as its real zeros?

Question

Elijah Allen · Accepted Answer

#5x^2-6x-4=0#

if #alpha " & "beta" are the roots of a quadratic equation"#

the equation can be written as

#x^2-(alpha+beta)x+alphabeta=0#

we have

#alpha=3/5+sqrt29/5#

#beta=3/5-sqrt29/5#

#alpha+beta=3/5+cancel(sqrt29/5)+3/5-cancel(sqrt29/5)#

#alpha+beta=6/5#

#alphabeta=(3/5+sqrt29/5)(3/5-sqrt29/5)#

#=9/25-29/25=-20/25=-4/5#

#:.x^2-(alpha+beta)x+alphabeta#

becomes

#x^2-6/5x-4/5=0#

#=>5x^2-6x-4=0#

Christian Antunez · Answer

#5x^2-6x-4# #"given the zeros ( roots) of a quadratic "x=a,x=b# #"then the factors are "(x-a),(x-b)# #"and the quadratic is the product of the factors"# #rArrp(x)=(x-a)(x-b)# #"here the roots are "x=3/5+sqrt29/5" and "x=3/5-sqrt29/5# #rArr"factors "(x-(3/5+sqrt29/5)),(x-(3/5-sqrt29/5))# #rArrp(x)=(x-3/5-sqrt29/5)(x-3/5+sqrt29/5)# #color(white)(rArrp(x))=(x-3/5)^2-(sqrt29/5)^2# #color(white)(rArrp(x))=x^2-6/5x+9/25-29/25# #color(white)(rArrp(x))=x^2-6/5x-4/5to(xx5)# #color(white)(rArrp(x))=5x^2-6x-4#

Sadie Ackerman · Answer

undefined To find a quadratic polynomial with integer coefficients having the given real zeros, we use the fact that if $ r_1 $ and $ r_2 $ are the roots of a quadratic polynomial $ ax^2 + bx + c $, then the polynomial can be expressed as $ a(x - r_1)(x - r_2) $.

Given the real zeros $ x_1 = \frac{3}{5} + \frac{\sqrt{29}}{5} $ and $ x_2 = \frac{3}{5} - \frac{\sqrt{29}}{5} $, we construct the quadratic polynomial using these roots.

The polynomial will have the form:
$$ (x - x_1)(x - x_2) $$

Substituting the given values of $ x_1 $ and $ x_2 $, we get:
$$ \left(x - \left(\frac{3}{5} + \frac{\sqrt{29}}{5}\right)\right)\left(x - \left(\frac{3}{5} - \frac{\sqrt{29}}{5}\right)\right) $$

$$ = \left(x - \frac{3}{5} - \frac{\sqrt{29}}{5}\right)\left(x - \frac{3}{5} + \frac{\sqrt{29}}{5}\right) $$

$$ = \left(x - \frac{3}{5}\right)^2 - \left(\frac{\sqrt{29}}{5}\right)^2 $$

$$ = \left(x - \frac{3}{5}\right)^2 - \frac{29}{25} $$

Expanding and simplifying, we get:
$$ x^2 - \frac{6}{5}x + \left(\frac{3}{5}\right)^2 - \frac{29}{25} $$

$$ = x^2 - \frac{6}{5}x + \frac{9}{25} - \frac{29}{25} $$

$$ = x^2 - \frac{6}{5}x - \frac{20}{25} $$

$$ = x^2 - \frac{6}{5}x - \frac{4}{5} $$

Therefore, the quadratic polynomial with integer coefficients having the given real zeros is $ x^2 - \frac{6}{5}x - \frac{4}{5} $.