Question

If the zeroes of the polynomial ax²+bx+c are in the ratio 2:5 proved that 10b²=49ac

Answer 1

Step-by-step explanation:

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Answer 2

Let the roots are 2x and 5x. Using the relation in coefficients and roots:

Sum of roots = -b/a = 2x + 5x = 7x

Product of roots = c/a = (2x)(5x) = 10x²

Therefore, left hand side of the question:

=> 10b²

=> 10b²/a² * a² {multiply & divide by a²}

=> 10(b/a)² * a²

=> 10(-7x)² * a² {b/a = -7x}

=> 49 (10x²) a²

=> 49 (c/a) a² {10x² = c/a}

=> 49ac

Hence, 10b² = 49ac.

Method 2: _______________

Roots = (-b ± √d)/2a, their ratio is 2:5

=> (-b - √d)/(-b + √d) = 2/5

=> -5b - 5√d = -2b + 2√d

=> - 3b = 7√d {square on both sides}

=> 9b² = 49(b² - 4ac) {d = b² - 4ac}

=> 9b² = 49b² - 49(4ac)

=> 4(49ac) = 40b²

=> 49ac = 10b² , proved.