Question

If one zero of the quadratic polynomial f(x)=4x^2-8kx-9
negative of the other, find the value of k
Answer with explanation
by calculating ​

Answer 1

Answer:

Given:

• $4 {x}^{2} - 8kx - 9$
• The zeroes are opposite in symbol.

Need to find:

• k=?

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$\huge{\underline{\underline{\red{SOLUTION}}}}$

let one zero be "p" then the other zero is "-p"

Now by relation between the zeroes of quadratic polynomial we know,

$\alpha + \beta = \dfrac{ - b}{a}$

and

$\alpha \beta = \dfrac{c}{a}$

when,

$\alpha \: and \: \beta \: are \: zeroes \: of \: polynomial$

and the equation is of the form:

$a {x}^{2} + bx + c$

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Here the zeroes are p and -p

a=4

b=(-8k)

c= (-9)

Thus,

$(p) \times ( - p) = \dfrac{ - 9}{4} \\ \implies - ( {p}^{2} ) = - ( \dfrac{9}{4} ) \\ \implies \: {p}^{2} = \dfrac{9}{4}$

$\pink{\boxed{\implies \: p = \pm \dfrac{3}{2}}}$

The zeroes are $\dfrac{3}{2}$ and $\dfrac{-3}{2}$

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now we know,

$sum \: of \: two \: zeros \: = \frac{ - b}{a} \\ \implies \: \frac{3}{2} + ( \frac{ - 3}{2}) = \frac{ - ( - 8k)}{4} \\ \implies \: \frac{3}{2} - \frac{3}{2} = \frac{8k}{4} \\ \implies \: 0 = 2k$

$\pink{\boxed{\implies \: k = 0}}$

Answer 2

Answer:

The question says one root is negative of other .

So, the sum of the root must be zeroe.

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Step-by-step explanation:

For any quadratic equation like: ax²+bx+c = 0

the sum of the root is given by -b/a

-b (cofficient of x)

a (cofficient of x²)

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So,in the equation = 4x²-8kx-9

b =-8k & a = 4

so, 0=-b/a

0 =-(-8k) /4

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$2k=0=> k=0$

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