Question

If a+b =6 and ab =-1 find the value of a and b

Answer 1

Answer:

Given,

ab = - 1

= > a = -1/b

According to question:

= > a + b = 6

= > (-1/b) + b = 6

= > (-1+b^2) = 6b

= > b^2 - 6b - 1 = 0

Using Quadratic formula-

$\implies b=\dfrac{6\pm\sqrt{6^2-4(-1)}}{2}\\\\\\\implies b=\dfrac{6\pm\sqrt{36+4}}{2}\\\\\\\implies b=\dfrac{6\pm\\sqrt{40}}{2}\\\\\\\implies b=\dfrac{6\pm 2\sqrt{10}}{2}\\\\\implies b = 3 \pm \sqrt{10}$

And therefore, a = 6 - b = 6 - 3 - $\sqrt{10}$

a = 3 - $\sqrt{10}$

Answer 2

Answer :

Given -

• a + b = 6 and ab = -1

To Find -

• Value of a and b ?

Solution -

We are Given, that, ab = -1.

So, a = -1/b.

Substituting this value,

⇒ a + b = 6

⇒ (-1/b) + b = 6

⇒ -1 + b² = 6b

⇒ b² - 6b - 1 = 0

Using Quadratic Formula :

⇒ $\sf{\dfrac{-b\pm \sqrt{{b}^{2} - 4ac}}{2a}}$

⇒ $\sf{\dfrac{-(-6)\pm \sqrt{{(-6)}^{2} - 4\times (1) \times (-1)}}{2\times (1)}}$

⇒ $\sf{\dfrac{6\pm \sqrt{36- (-4)}}{2}}$

⇒ $\sf{\dfrac{6\pm \sqrt{36+4}}{2}}$

⇒ $\sf{\dfrac{6\pm \sqrt{40}}{2}}$

⇒ $\sf{\dfrac{6\pm 2\sqrt{10}}{2}}$

⇒ $\sf{\dfrac{2(3\pm \sqrt{10})}{2}}$

⇒ $\sf{3\pm\sqrt{10}}$

Hence, b = $\bf\underline{3+ \sqrt{10}}$ OR $\bf\underline{3- \sqrt{10}}$

Now, Finding b.

We know that, a = -1/b.

So, a = $\bf\underline{\dfrac{-1}{3+\sqrt{10}}}$ OR $\bf\underline{\dfrac{-1}{3- \sqrt{10}}}$