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?

Mina Whitehead
2022-04-11
Answered

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

Answered 2022-04-12
Author has **15** answers

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,

${x}^{2}+ax+\frac{{a}^{2}}{4}={(x+\frac{a}{2})}^{2}=0$

Can you work things out from here

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Can you work things out from here

Raina Blackburn

Answered 2022-04-13
Author has **10** answers

Step 1

${x}^{2}+ax+b$

$\iff {(2x+a)}^{2}={a}^{2}-4b$

If the quadratic equation has one real root, then

$\mathrm{\Delta}={a}^{2}-4b=0$

This implies,

$\{\begin{array}{l}2x+a=0\\ {a}^{2}=4b\end{array}{\textstyle \phantom{\rule{0.278em}{0ex}}}\u27f9{\textstyle \phantom{\rule{0.278em}{0ex}}}\{\begin{array}{l}a=-2x\\ b={x}^{2}\end{array}$

Step 2

If the quadratic has one real root, then,$x}_{1}={x}_{2$

By the Vieta's formulas, we have

$\{\begin{array}{l}{x}_{1}+{x}_{2}=-a\\ {x}_{1}{x}_{2}=b\end{array}{\textstyle \phantom{\rule{0.278em}{0ex}}}\u27f9{\textstyle \phantom{\rule{0.278em}{0ex}}}\{\begin{array}{l}a=-2x\\ b={x}^{2}\end{array}$

Step 3

If the quadratic has one real root, then

$x}^{2}+ax+b={(x-{x}_{1})}^{2$

$\Rightarrow {x}^{2}+ax+b={x}^{2}-2{\times}_{1}+{x}_{1}^{2}$

$\Rightarrow a=-2{x}_{1},\text{}b={x}_{1}^{2}$

If the quadratic equation has one real root, then

This implies,

Step 2

If the quadratic has one real root, then,

By the Vieta's formulas, we have

Step 3

If the quadratic has one real root, then

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