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If the product of the zeroes of the quadratic polynomial p(x)=ax^(2)-6x-6 is 4 , then find the value of a. Also, find the sum of the zeroes of the polynomial.

If the product of the zeroes of the quadratic polynomial $p(x)=a x^{2}-6 x-6$ is $4$ , then find the value of $a$ . Also, find the sum of the zeroes of the polynomial.

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Conjugate root theorems

Full solution

Q. If the product of the zeroes of the quadratic polynomial

p(x)=a x^{2}-6 x-6

is

4

, then find the value of

a

. Also, find the sum of the zeroes of the polynomial.

Denote the zeroes: Let's denote the zeroes of the polynomial $p(x) = ax^2 - 6x - 6$ as $\alpha$ (alpha) and $\beta$ (beta). According to Vieta's formulas, for a quadratic polynomial $ax^2 + bx + c$ , the product of the roots is given by $\frac{c}{a}$ and the sum of the roots is given by $-\frac{b}{a}$ .
Use Vieta's formula: Given that the product of the zeroes $(\alpha\beta)$ is $4$ , we can write the equation using Vieta's formula: $\alpha\beta = \frac{c}{a}$ . In our case, $c = -6$ and we are given $\alpha\beta = 4$ . So we have $-\frac{6}{a} = 4$ .
Solve for a: Solving for a, we multiply both sides by $a$ to get $-6 = 4a$ . Then, we divide both sides by $4$ to find the value of $a$ : $a = -6/4 = -3/2$ .
Find sum of zeroes: Now, we need to find the sum of the zeroes $(\alpha + \beta)$ . Using Vieta's formula again, we have $\alpha + \beta = -\frac{b}{a}$ . In our case, $b = -6$ and we have already found $a = -\frac{3}{2}$ . So we have $\alpha + \beta = -(-6)/(-\frac{3}{2})$ .
Simplify the expression: Simplifying the expression for the sum of the zeroes, we get $\alpha + \beta = \frac{6}{(-\frac{3}{2})} = 6 \times (-\frac{2}{3}) = -4$ .

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