Having trouble with this problem:

## The two solutions of the quadratic equation 3x^2-x+k are p/4 and p+1. Find the values of p and k.

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# 1 Answer

Using the relations between roots and coefficients:

For ax^2+bx+c = 0, the two roots x1 and x2 have the following relations:

x1+x2 = -b/a

x1*x2 = c/a

Therefore, we have

(p/4) + (p+1) = 1/3 ......(1) => p = -8/15

(p/4)(p+1) = k/3 ......(2)

Plug p = -8/15 into (2), and solve for k,

k = -14/75

Check:

3(x - p/4)(x - p - 1) = 3(x + 2/15)(x - 7/15) = 3x^2 - x - 14/75

## Comments

Thanks but can you show me your working for:

(p/4) + (p+1) = 1/3 and "Plug p = -8/15 into (2), and solve for k" I don't understand how you have done these

(p/4) + (p+1) = 1/3

Multiply both sides by 12,

3p+12p+12 = 4 => 15p = -8 => p = -8/15

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(p/4)(p+1) = k/3 => k = (3/4)p(p+1) = (3/4)(-8/15)(7/15) = -14/75

Thank you

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