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If both roots of x²−px−q=0 are positive and one root is the cube of another, then:

a. p>0,q>0
b. p<0,q>0
c. p>0,q<0
d. p<0,q<0

User Jigar Tank
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1 Answer

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Final answer:

Using Vieta's formulas, when a quadratic equation has two positive roots, p equals the sum and q equals the product of the roots; since one root is the cube of the other, both p and q are positive. Therefore, p > 0 and q = -rs > 0. So the correct answer is option a. p > 0, q > 0.

Step-by-step explanation:

If both roots of the quadratic equation x²-px-q=0 are positive and one root is the cube of another, then we can determine the signs of p and q. Let's assume the two roots are a and a³, where a is the smaller root and a³ is the larger root. By Vieta's formulas, p is the sum of the roots and q is the product of the roots of the quadratic equation.

For the given equation:

  • The sum of the roots (p) is a + a³. Since both a and a³ are positive, their sum would also be positive, thus p > 0.
  • The product of the roots (q) is a × a³ = a⁴. Since a is positive, a⁴ would also be positive, thus q > 0.

The quadratic equation x² - px - q = 0 has two roots, let's call them r and s. We are given that both roots are positive and that one root is the cube of the other. So we can write the equation as:

(x - r)(x - s) = x² - (r + s)x + rs = 0

We know that r > 0 and s = r³. To find the values of p and q, we can compare the coefficients of the equation with the expression (r + s) and rs:

From the quadratic equation, we have:

r + s = p (1)

rs = -q (2)

Since both roots are positive, r > 0 and s = r³ > 0. Therefore, p > 0 and q = -rs > 0. So the correct answer is option a. p > 0, q > 0.

User Goodm
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