The x2+ax+b=0 has only one solution and this is x=−1+12. What are the values of a and b?

Question: "The x2+ax+b=0 has only one solution and this..."

High School
Algebra
Algebra I
Quadratic function and equation

Mina Whitehead

Answered question

2022-04-11

The

x^{2} + a x + b = 0

has only one solution and this is

x = - 1 + \frac{1}{2}

. What are the values of a and b?

Answer & Explanation

tutaonana223a

Beginner2022-04-12Added 15 answers

Step 1
If

x^{2} + a x + 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 = a^{2} - 4 b = 0

and so

4 b = a^{2}

. Hence,

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

Can you work things out from here

Raina Blackburn

Beginner2022-04-13Added 10 answers

Step 1

x^{2} + a x + b

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

If the quadratic equation has one real root, then

Δ = a^{2} - 4 b = 0

This implies,

{\begin{cases} 2 x + a = 0 \\ a^{2} = 4 b \end{cases} ⟹ {\begin{cases} a = - 2 x \\ b = x^{2} \end{cases}

Step 2
If the quadratic has one real root, then,

x_{1} = x_{2}

By the Vieta's formulas, we have

{\begin{cases} x_{1} + x_{2} = - a \\ x_{1} x_{2} = b \end{cases} ⟹ {\begin{cases} a = - 2 x \\ b = x^{2} \end{cases}

Step 3
If the quadratic has one real root, then

x^{2} + a x + b = {(x - x_{1})}^{2}

\Rightarrow x^{2} + a x + b = x^{2} - 2 \times_{1} + x_{1}^{2}

\Rightarrow a = - 2 x_{1}, b = x_{1}^{2}

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