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

Mina Whitehead 2022-04-11 Answered
The x2+ax+b=0 has only one solution and this is x=1+12. What are the values of a and b?
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Answers (2)

tutaonana223a
Answered 2022-04-12 Author has 15 answers
Step 1
If x2+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 Δ must equal 0. Here, this means that
Δ=a24b=0
and so 4b=a2. Hence,
x2+ax+a24=(x+a2)2=0
Can you work things out from here
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Raina Blackburn
Answered 2022-04-13 Author has 10 answers
Step 1
x2+ax+b
(2x+a)2=a24b
If the quadratic equation has one real root, then
Δ=a24b=0
This implies,
{2x+a=0a2=4b{a=2xb=x2
Step 2
If the quadratic has one real root, then, x1=x2
By the Vieta's formulas, we have
{x1+x2=ax1x2=b{a=2xb=x2
Step 3
If the quadratic has one real root, then
x2+ax+b=(xx1)2
x2+ax+b=x22×1+x12
a=2x1, b=x12
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