# The $$\displaystyle{x}^{{{2}}}+{a}{x}+{b}={0}$$ has only one solution and

The ${x}^{2}+ax+b=0$ has only one solution and this is $x=-1+\frac{1}{2}$. What are the values of a and b?
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tutaonana223a
Step 1
If ${x}^{2}+ax+b=0$ has one solution, then the quadratic has a repeated root. Both of the roots could be equal to 5, for instance. If the roots are repeated, then the discriminant $\mathrm{\Delta }$ must equal 0. Here, this means that
$\mathrm{\Delta }={a}^{2}-4b=0$
and so $4b={a}^{2}$. Hence,
${x}^{2}+ax+\frac{{a}^{2}}{4}={\left(x+\frac{a}{2}\right)}^{2}=0$
Can you work things out from here
###### Not exactly what you’re looking for?
Raina Blackburn
Step 1
${x}^{2}+ax+b$
$⇔{\left(2x+a\right)}^{2}={a}^{2}-4b$
If the quadratic equation has one real root, then
$\mathrm{\Delta }={a}^{2}-4b=0$
This implies,
$\left\{\begin{array}{l}2x+a=0\\ {a}^{2}=4b\end{array}\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\left\{\begin{array}{l}a=-2x\\ b={x}^{2}\end{array}$
Step 2
If the quadratic has one real root, then, ${x}_{1}={x}_{2}$
By the Vieta's formulas, we have
$\left\{\begin{array}{l}{x}_{1}+{x}_{2}=-a\\ {x}_{1}{x}_{2}=b\end{array}\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\left\{\begin{array}{l}a=-2x\\ b={x}^{2}\end{array}$
Step 3
If the quadratic has one real root, then
${x}^{2}+ax+b={\left(x-{x}_{1}\right)}^{2}$
$⇒{x}^{2}+ax+b={x}^{2}-2{×}_{1}+{x}_{1}^{2}$