Form the quadratic equation whose roots $α$ $alpha$ and $β$ $beta$ satisfy the relations $α β = 768$ $alpha beta=768$ and $α^{2} + β^{2} = 1600$ $alpha^2+beta^2=1600$ ?

Question

17135 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T09:54:40

$x^{2} - 56 x + 768 = 0$ $x^2-56x+768=0$

if $α and β$ $alpha " and " beta$ are the roots of a quadratic eqn , the eqn can be written as

$x^{2} - (α + β) x + α β = 0$ $x^2-(alpha+beta)x+alphabeta=0$

we are given

$α β = 768$ $color(red)(alpha beta=768)$

$α^{2} + β^{2} = 1600$ $color(blue)(alpha^2+beta^2=1600)$

now
${(α + β)}^{2} = α^{2} + β^{2} + 2 α β$ $(alpha+beta)^2=color(blue)(alpha^2+beta^2)+color(red)(2alphabeta)$

so ${(α + β)}^{2} = 1600 + 2 \times 768 = 3136$ $(alpha+beta)^2=color(blue)(1600)+2xxcolor(red)(768)=3136$

$:.alpha+beta=sqrt(3136)=56$

the required quadratic is then

$x^2-56x+768=0$

Form the quadratic equation whose roots alphaαalpha and betaβbeta satisfy the relations alpha beta=768αβ=768alpha beta=768 and alpha^2+beta^2=1600α2+β2=1600alpha^2+beta^2=1600?