Question

if a and b are roots of the quadratic equation x^2-x+1=0, the write the value of a^2+b^2​

Answer 1

Answer:

a²+b²= -1

Step-by-step explanation:

if a and b are roots of the quadratic equation

x²-x+1=0

Then a+b=-(-1/1)=1

ab=1/1=1

Now a+b=1

squaring we get

(a+b)²=1

a²+b²+2ab=1

a²+b² +2*1=1

a²+b²=1-2=-1

Answer 2

Answer:

$a^{2}+b^{2} = -1$

Step-by-step explanation:

Given, $x^{2}-x+1 = 0$

⇒ $D = b^{2}-4ac$

⇒ $D = 1-4 = -3$

⇒ $x = \frac{-b+\sqrt{D}}{2a}$ or $x = \frac{-b-\sqrt{D}}{2a}$

⇒ $x = \frac{1+\sqrt{-3}}{2}$ or $x = \frac{1-\sqrt{-3}}{2}$

⇒ $x = \frac{1+\sqrt{3}i}{2}$ or $x = \frac{1-\sqrt{3}i}{2}$

⇒ $a = \frac{1}{2}+\frac{\sqrt{3}i}{2}$ and $b = \frac{1}{2}-\frac{\sqrt{3}i}{2}$

⇒ $a^{2} = (\frac{1}{2}+\frac{\sqrt{3}i}{2})^{2}$ and $b^{2} = (\frac{1}{2}-\frac{\sqrt{3}i}{2})^{2}$

⇒ $a^{2} = -\frac{1}{2}+\frac{\sqrt{3}i}{2}$ and $b^{2} = -\frac{1}{2}-\frac{\sqrt{3}i}{2}$

⇒ $a^{2}+b^{2} = -\frac{1}{2}+\frac{\sqrt{3}i}{2} -\frac{1}{2}-\frac{\sqrt{3}i}{2}$

⇒ $a^{2}+b^{2} = -\frac{2}{2}$

⇒ $a^{2}+b^{2} = -1$