Question

5. If one of the zeroes of the quadratic polynomial (2k – 1)x2 + 3kx + lis - 5, then the value of k is
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Answer 1

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Answer 2

Step-by-step explanation:

Given: $(2k - 1) {x}^{2} + 3kx + 1$

To find: If one of the zeroes of the given quadratic polynomial - 5, then the value of k is ?

Solution:

Let the polynomial is p(x) and if -5 is one of the zero of then p(-5)=0.

Put the value of zero in the polynomial.

$p( - 5) = (2k - 1)( { - 5)}^{2} + 3k( - 5) + 1$

$25(2k - 1) - 15k + 1 = 0$

$50k - 25 - 15k + 1 = 0$

$35k = 24$

$\bf k=\frac{24}{35}$

Final answer:

If one of the zeroes of the

$(2k - 1) {x}^{2} + 3kx + 1$ quadratic polynomial - 5, then the value of k is $\frac{24}{35}$

Hope it will help you.

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