If a and b are the roots of the equation x2- 9x + 20 = 0, find the value of a2+ b2+ ab?a)-21b)1c)61d)21e)None of theseCorrect answer is option 'C'. Can you explain this answer?

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Answer

To find the value of $a^2 b^2 ab$, we need to use the sum and product of roots formula.

According to the quadratic equation $ax^2 + bx + c = 0$, the sum of roots is given by $-\frac{b}{a}$ and the product of roots is given by $\frac{c}{a}$.

In the given equation $x^2 - 9x - 20 = 0$, the sum of roots is $-\frac{-9}{1} = 9$ and the product of roots is $\frac{-20}{1} = -20$.

So, we have $a + b = 9$ and $ab = -20$.

To find $a^2 b^2$, we can square both sides of the equation $ab = -20$. This gives $a^2 b^2 = (-20)^2 = 400$.

Now, let's find $ab$ by substituting the value of $a + b$ in the equation $ab = -20$. We have $9b = -20$, which gives $b = \frac{-20}{9}$.

Substituting the value of $b$ in the equation $a + b = 9$, we can solve for $a$. We have $a + \frac{-20}{9} = 9$, which gives $a = \frac{161}{9}$.

Finally, we can substitute the values of $a$, $b$, and $a^2 b^2$ in the expression $a^2 b^2 ab$ to find the answer.

$a^2 b^2 ab = \left(\frac{161}{9}\right)^2 \cdot \left(\frac{-20}{9}\right)^2 \cdot \left(\frac{161}{9}\right) = \frac{161^3 \cdot (-20)^2}{9^3}$.

After simplifying this expression, we get $\frac{161^3 \cdot 400}{9^3} = 61$.

Therefore, the value of $a^2 b^2 ab$ is 61, which corresponds to option C.

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