zeroes of $9 x^{2} - 24 x + 16$

both positive and unequal

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both negative and unequal

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one negative and one positive

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both equal

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Solution

The correct option is D
both equal

X coefficient is negative, constant is positive $\to$ both zeroes are +ve and equal

Since zeroes are $\frac{4}{3}, \frac{4}{3}$

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Similar questions

{9 x}^{2} + 24 x + 16

(9 x^{2} - 24 x + 16)

(3 x - 4)

x^{3} - 9 x^{2} + 24 x - 16 = 0

Find the value of the expression
$(9 x^{2} + 24 x + 16)$ when $x = 12$

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