Question

if product of the zeroes of the polynomial kx square+41x+42 is 7 then find the zeroes of the polynomial (k-4)x square+(k+1)x+5

Answer 1

hope this answer helps

Answer 2

Concept:

The sum of zeros of a quadratic expression is equal to the negative value of the ratio of coefficient of variable with power 1 and coefficient of variable with power 2.

The Product of zeros of a quadratic expression is equal to the value of the ratio of constant term and coefficient of variable with power 2.

For example if the quadratic expression is $ax^2+bx+c$ and the zeros of this equation are m, n then,

m + n = -b/a

mn = c/a

Given:

Given that, product of the zeroes of the polynomial $kx^2+41x+42$ is 7.

Find:

The zeroes of the polynomial $(k-4)x^2+(k+1)x+5$.

Solution:

Given the first quadratic polynomial is $kx^2+41x+42$.

So here a = k, b = 41 and c = 42

Also it is given that the product of zeros is 7.

According to quadratic polynomial rule,

c/a = 7

42/k = 7

k = 42/7

k = 6

Thus the value of k = 6

Now the another polynomial becomes,

 $(k-4)x^2+(k+1)x+5\\=(6-4)x^2+(6+1)x+5\\=2x^2+7x+5$

So the zeros we can find by solving

 $2x^2+7x+5=0\\2x^2+2x+5x+5=0\\2x(x+1)+5(x+1)=0\\(2x+5)(x+1)=0$

Since if the product of two number is zero then at least one of them is zero.

So, either,

2x+5 = 0

2x = -5

x = -5/2

Or,

x+1 = 0

x = -1

Hence the zeros of the polynomial $\mathbf{kx^2+41x+42}$ are -5/2 and -1.

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