Question

Hi. I need help with these questions.
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Answer 16 and 17

Answer 1

Explanation:

16.................................................

Given

• Equation ax² + bx + c = 0
• One of the roots is twice the other

To find

• Show that 2b² = 9ac

Solution

Let the roots be p and q

The sum and product of the roots:

• p + q = -b/a
• pq = c/a

The sum of the roots, considering p = 2q,

• p + q = 2q + q = 3q

And

• 3q = -b/a ⇒ q = -b/(3a)

The product of the roots, considering p = 2q:

• pq = 2q(q) = 2q²

And

• 2q² = c/a

Substitute q with -b/(3a)

• 2(-b/(3a))² = c/a
• 2b²/(9a²) = c/a
• 2b² = 9a²(c/a)
• 2b² = 9ac

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17. .................................................

Given

Quadratic equation

• ax² = bx + c = 0, with equal roots

To find

• Show that b² = 4ac

Solution

Equal roots means the discriminant is zero:

• D = √b² - 4ac
• √b² - 4ac = 0
• b² - 4ac = 0
• b² = 4ac