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.

Find Zeroes of Polynomial: Let's denote the zeroes of the polynomial p(x) = ax^2 - 6x - 6 as α (alpha) and β (beta). According to Vieta's formulas, the product of the zeroes of a quadratic polynomial ax^2 + bx + c is c/a. We are given that the product of the zeroes is 4. So, we have: αβ = c/a Given that c = -6 and αβ = 4, we can write: -6/a = 4 Now, we solve for a: a = -6/4 a = -3/2Calculate Product of Zeroes: Next, we need to find the sum of the zeroes of the polynomial. According to Vieta's formulas, the sum of the zeroes of a quadratic polynomial ax^2 + bx + c is -b/a. We have: α + β = -b/a Given that b = -6 and a = -3/2 (from the previous step), we can write: α + β = -(-6)/(-3/2) α + β = 6/(-3/2) α + β = 6 * (-2/3) α + β = -4

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

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

4

, then find the value of

a

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

Find Zeroes of Polynomial: 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, the product of the zeroes of a quadratic polynomial $ax^2 + bx + c$ is $\frac{c}{a}$ . We are given that the product of the zeroes is $4$ . So, we have: $\alpha\beta = \frac{c}{a}$ Given that $c = -6$ and $\alpha\beta = 4$ , we can write: $-\frac{6}{a} = 4$ Now, we solve for $\alpha$ $0$ : $\alpha$ $1$ $\alpha$ $2$
Calculate Product of Zeroes: Next, we need to find the sum of the zeroes of the polynomial. According to Vieta's formulas, the sum of the zeroes of a quadratic polynomial $ax^2 + bx + c$ is $-b/a$ . We have: $\newline$ $\alpha + \beta = -b/a$ $\newline$ Given that $b = -6$ and $a = -3/2$ (from the previous step), we can write: $\newline$ $\alpha + \beta = -(-6)/(-3/2)$ $\newline$ $\alpha + \beta = 6/(-3/2)$ $\newline$ $\alpha + \beta = 6 \times (-2/3)$ $\newline$ $\alpha + \beta = -4$