a) α? + β2 +αβ= 0 give solution
Answers
Answer:
α=
2a
−b+
b
2
−4ac
and
\displaystyle\beta=\frac{{-{b}-\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}β=
2a
−b−
b
2
−4ac
Sum of the roots α and β
We can add \displaystyle\alphaα and \displaystyle\betaβ as follows:
\displaystyle\alpha+\beta=\frac{{-{b}+\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}+\frac{{-{b}-\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}α+β=
2a
−b+
b
2
−4ac
+
2a
−b−
b
2
−4ac
\displaystyle=\frac{{-{2}{b}+{0}}}{{{2}{a}}}=
2a
−2b+0
\displaystyle=-\frac{b}{{a}}=−
a
b
We can multiply \displaystyle\alphaα and \displaystyle\betaβ as follows. First, recall that in general,
\displaystyle{\left({X}+{Y}\right)}{\left({X}-{Y}\right)}={X}^{2}-{Y}^{2}(X+Y)(X−Y)=X
2
−Y
2
and
\displaystyle{\left(\sqrt{{{X}}}\right)}^{2}={X}(
X
)
2
=X
We make use of these to obtain:
\displaystyle\alpha\times\beta=\frac{{-{b}+\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}\times\frac{{-{b}-\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}α×β=
2a
−b+
b
2
−4ac
×
2a
−b−
b
2
−4ac
\displaystyle=\frac{{{\left(-{b}\right)}^{2}-{\left(\sqrt{{{b}^{2}-{4}{a}{c}}}\right)}^{2}}}{{\left({2}{a}\right)}^{2}}=
(2a)
2
(−b)
2
−(
b
2
−4ac
)
2
\displaystyle=\frac{{{b}^{2}-{\left({b}^{2}-{4}{a}{c}\right)}}}{{{4}{a}^{2}}}=
4a
2
b
2
−(b
2
−4ac)
\displaystyle=\frac{{{4}{a}{c}}}{{{4}{a}^{2}}}=
4a
2
4ac
\displaystyle=\frac{c}{{a}}=
a
c
Summary
The sum of the roots \displaystyle\alphaα and \displaystyle\betaβ of a quadratic equation are:
\displaystyle\alpha+\beta=-\frac{b}{{a}}α+β=−
a
b
The product of the roots \displaystyle\alphaα and \displaystyle\betaβ is given by:
\displaystyle\alpha\beta=\frac{c}{{a}}αβ=
a
c