Question 767787
Q:
Find the quadratic equation whose roots are twice the roots of {{{4x^2+8x-5=0}}}
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In {{{ax^2 + bx + c = 0}}}, the sum of the roots is {{{(-b)/a}}} and the product of the roots is {{{c/a}}}.
Let r and s be the roots of {{{4x^2+8x-5=0}}}.
So the sum of the roots is r + s = {{{(-8)/4}}} = -2 and the product of the roots is rs = {{{(-5)/4}}}.
2r + 2s = 2(r + s) = 2(-2) = -4
(2r)(2s) = 4(rs) = {{{4((-5)/4)}}} = -5
The new quadratic equation is:
{{{x^2}}} - (sum of roots)x + (product of roots) = 0
Answer: {{{highlight(x^2 + 4x - 5 = 0)}}}