If 64x^(2)+px+81=(8x+q)^(2), where p and q are positive constants, what is the value of (p)/(q) ?

Question

If  64x^(2)+px+81=(8x+q)^(2), where  p and  q are positive constants, what is the value of  (p)/(q) ?

Bytelearn · Accepted Answer

Expand equation: Expand the right side of the equation (8x+q)^(2) to get 64x^2 + 16qx + q^2.Compare coefficients: Compare the coefficients of the corresponding terms on both sides of the equation. We have 64x^2 on both sides, so they match. Now, compare the linear term and the constant term. We have px on the left side and 16qx on the right side, so p = 16q. For the constant terms, we have 81 on the left side and q^2 on the right side, so q^2 = 81.Find q: Since q^2 = 81 and q is a positive constant, we take the positive square root of 81 to find q. Therefore, q = 9.Find p: Substitute q = 9 into the equation p = 16q to find the value of p. So, p = 16 * 9.Calculate p/q: Calculate the value of p using the value of q. Therefore, p = 16 * 9 = 144.Simplify p/q: Now that we have both p and q, we can find the value of (p)/(q). So, (p)/(q) = 144/9.Simplify the fraction 144/9 to get the final value of (p)/(q). Therefore, (p)/(q) = 16.

If $64 x^{2}+p x+81=(8 x+q)^{2}$ , where $p$ and $q$ are positive constants, what is the value of $\frac{p}{q}$ ?

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If 64x2+px+81=(8x+q)2 64 x^{2}+p x+81=(8 x+q)^{2} 64x2+px+81=(8x+q)2, where p p p and q q q are positive constants, what is the value of pq \frac{p}{q} qp​ ?

If $64 x^{2}+p x+81=(8 x+q)^{2}$ , where $p$ and $q$ are positive constants, what is the value of $\frac{p}{q}$ ?