If the roots of ax^2- 4x + b = 0 are 1/2 and 8, what is the value of (a + b)^2?

2 Answers | Add Yours

Top Answer

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

Let’s start with the given equation:

ax^2 - 4x + b =0

we write it as

=> x^2 - (4/a)x + (b/a) =0

now as -1/2 and -8 are the roots of the equation we have:

=> (x-1/2)(x-8) = x^2 - (4/a)x + (b/a)

=> x^2 –(1/2 + 8)x + 4 = x^2 - (4/a)x + (b/a)

So we have  8+ 1/2 = 4/a and 4 = b/a

8+ 1/2 = 4/a

=> a = 8/17

b = 4a = 32/17

Therefore (a + b)^2 = (8/17 + 32/17)^2 = (40/17)^2 = 1600/ 289

The value of  (a+b)^2 = 1600/ 289

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

ax^2-4x+b = 0 has the roots 1/2 and 8.

Therefore by the relations between  the roots and coefficients, we have:

sum of the roots: 

1/2+8 = -(-4)/a = 4/a.

8.5 = 4/a

a = 4/8.5 = 8/17.

Product of the roots:

(1/2)*8 = b/a

4 = b/a

Therefore b = 4*a = 4*8/17 = 32/17

Thus a = 8/17 and b = 32/17. Put these values of a and b in (a+b)^2.

Therefore (a+b)^2 = (8/17 +32/17)^2

(a+b)^2 = (40/17)^2.

(a+b)^2 =  1600/289.

We’ve answered 318,955 questions. We can answer yours, too.

Ask a question