If the product of the zeros of p(x) = ax^{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For a quadratic polynomial $p(x) = ax^{2} + bx + c$,the product of the zeros is given by the formula $\frac{c}{a}$.Given the polynomial $p(x) = ax

Vedclass · Accepted Answer

$-3/2$

Vedclass · Answer

(C) For a quadratic polynomial $p(x) = ax^{2} + bx + c$,the product of the zeros is given by the formula $\frac{c}{a}$.Given the polynomial $p(x) = ax^{2} - 6x - 6$,we identify the coefficients as $A = a$,$B = -6$,and $C = -6$.The product of the zeros is given as $4$.Therefore,$\frac{C}{A} = 4$.Substituting the values,we get $\frac{-6}{a} = 4$.Solving for $a$: $4a = -6$,which implies $a = -\frac{6}{4} = -\frac{3}{2}$.Thus,the correct value of $a$ is $-\frac{3}{2}$.

If the product of the zeros of $p(x) = ax^{2} - 6x - 6$ is $4$,then $a = $ ............

Similar Questions

The product of two polynomials is $x^{2}-x-72$ and if one of the polynomials is $(x+8)$,then the other polynomial is $\ldots \ldots \ldots \ldots . .$

If $\frac{1}{3}$ is a zero of $p(x) = 3x^2 - 10x + 3$,then the other zero is ...........

If the sum of the zeros is $-3$ and the product of the zeros is $-4,$ then the quadratic polynomial is $\ldots \ldots \ldots$

There are $x^{2}-2$ students in the class. $x^{3}-3x^{2}+5x-3$ chocolates are distributed between them. Each student should get the maximum possible number of chocolates. Find the number of chocolates received by each student and the number of chocolates left undistributed $(x \in N)$.

What must be added to the polynomial $p(x) = 6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ so that the resulting polynomial is exactly divisible by $3x^2 - 2x + 4$? (Hint: add $-r(x)$ to $p(x)$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module