Math, asked by avni07singh, 8 months ago

if one zero of f(x)=4x²-8kx+8x-9 is negative of the other, them find the zeros of kx²+kx+2​

Answers

Answered by yashkumar1939
1

Step-by-step explanation:

The zeroes are -1 and -2

Step-by-step explanation:

\text{Given that one zero of quadratic polynomial }4x^2-8kx+8x-9 \text{ is negative of other then}Given that one zero of quadratic polynomial 4x

2

−8kx+8x−9 is negative of other then

\text{we have to find the zeroes of }kx^2+3kx+2we have to find the zeroes of kx

2

+3kx+2

\text{ As one zero of quadratic polynomial }4x^2-8kx+8x-9 \text{ is negative of other } As one zero of quadratic polynomial 4x

2

−8kx+8x−9 is negative of other

∴ let the zeroes are α and -α

\text{sum of roots= }\alpha+(-\alpha)=\frac{-b}{a}=\frac{-(-8k+8)}{4}sum of roots= α+(−α)=

a

−b

=

4

−(−8k+8)

⇒ \frac{-(-8k+8)}{4}=0

4

−(−8k+8)

=0

⇒ \frac{8k-8}{4}=0

4

8k−8

=0

8k-8=08k−8=0

k=1k=1

The polynomial becomes

(1)x^2+3(1)x+2(1)x

2

+3(1)x+2

x^2+3x+2x

2

+3x+2

x^2+2x+x+2x

2

+2x+x+2

x(x+2)+1(x+2)x(x+2)+1(x+2)

(x+1)(x+2)(x+1)(x+2)

x+1=0 ⇒ x=-1

x+2=0 ⇒ x=-2

Hence, the zeroes are -1 and -2

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Answered by bhartipustakbhandar
0

Answer:

According to the given question it is already given that - ''If one zero of the quadratic polynomial f(x)=4 x² - 8 k x + 8 x - 9 is negative''

So let the roots of this polynomial be as follows :-

⇒ α , - α { As per to given )

We already know what is the sums of roots , ( i.e ...)

\alpha + ( - \alpha ) =α+(−α)= coefficient of x / coefficient of x²

\alpha - \alpha = \frac{-8k+8}{4}α−α=

4

−8k+8

→ ( As per to given )

0 = -2k + 20=−2k+2 → ( After simplifying )

2k-22k−2

k = \frac{2}{2} = 1k=

2

2

=1

Hence the value of k is 1

----------------------------------------------------------------------------------------------

⇒ Now let us put the value of k as 1 in the given polynomial - kx² + 3 kx + 2

⇒ So we get as follows ,

⇒1 x^{2} + 3 x + 21x

2

+3x+2

→ Now , we can find the zeroes of polynomial by spliting the middle term method for 1 x^{2} + 3 x + 21x

2

+3x+2 -

x^{2} + 3x + x = 0x

2

+3x+x=0

x^{2} + 2x+ x+ 2 = 0x

2

+2x+x+2=0

x ( x + 2 ) + 1 ( x + 2 ) = 0x(x+2)+1(x+2)=0

(x + 1 ) ( x+ 2) = 0(x+1)(x+2)=0

→ Now , let us find the alpha and beta for the following polynomial to get the final answer :-

x + 1 = 0x+1=0

x = - 1x=−1

→Alpha - α = -1

----------------------------------------------------------------------------------------------

Another Zero :-

x + 2 = 0x+2=0

x = -2x=−2

→Beta - β = -2

Hence the two zeroes are -1 and -2.

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