Question

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And give proper answers​

Answer 1

Step-by-step explanation:

Given :-

The ratio of the two roots of ax²+bx+c = 0 is 1:r and a≠0

To find :-

Show that (r+1)²/r = b²/ac.

Solution :-

Given quadratic equation is ax²+bx+c = 0

a ≠ 0

Let P(x) = ax²+bx+c = 0

Let the roots of P(x) be A and B

We know that

Sum of the roots = -b/a

=> A + B = -b/a --------(1)

Product of the roots = c/a

=> A×B = c/a -------(2)

The ratio of the roots = A:B

According to the given problem

Ratio of the roots of P(x) = 1:r

=> A:B = 1:r

=> A/B = 1/r

=> 1/r = A/B

=> r = B/A ----------(3)

Now,

On taking LHS = (r+1)²/r

=> [(B/A)+1]²/(B/A)

=> [(B+A)/A]²/(B/A)

=> [(A+B)/A]²/(B/A)

=> [(A+B)²/A²]/(B/A)

=> [(A+B)²/A²]×(A/B)

=> [(A+B)²×A]/[(A²×B)]

=> (A+B)²/(AB)

From (1)&(2)

=> (-b/a)²/(c/a)

=> (b²/a²)/(c/a)

=> (b²/a²)×(a/c)

=> (b²×a)/(a²×c)

=> b²/ac

=> RHS

=> LHS = RHS

Hence, Proved.

Answer:-

If The ratio of the two roots of ax²+bx+c = 0 is 1:r and a≠0 then (r+1)²/r = b²/ac.

Used formulae:-

• The standard quadratic equation is ax²+bx+c = 0
• Sum of the roots = -b/a
• Product of the roots = c/a