If the 2 real number solutions of x^2 - 7x + 4 are represented by p

Question

If the 2 real number solutions of  x^2 - 7x + 4 = 2  are represented by p and q, what is |p - q| ?

A. 1

B. 5

C. 7

D.  \sqrt{33}

E.  \sqrt{41}

gmatophobia · Accepted Answer

Roots of equation represented by  ax^2 + bx + c  are  \frac{-b\pm \sqrt{b^2-4ac}}{2a}

x^2 - 7x + 4 = 2

x^2 - 7x + 4 - 2 = 0

x^2 - 7x + 2 = 0

Option 1   : Solving the roots

Roots =  \frac{7 \pm \sqrt{49-8}}{2}

p =  \frac{7 + \sqrt{41}}{2}

q =  \frac{7 - \sqrt{41}}{2}

|p-q| =  \frac{7 + \sqrt{41}}{2} - \frac{7 + \sqrt{41}}{2}

|p-q| =  \frac{\sqrt{41}}{2} + \frac{\sqrt{41}}{2}

|p-q| =  \sqrt{41}

Option 2   : Using Formula

The difference in roots is given by  \frac{\sqrt{b^2 - 4ac}}{a}

|p-q| =  \frac{\sqrt{49 - 8}}{1} = \sqrt{41}

Option E

KarishmaB · Answer

Responding to a pm: Can we do it without the formula? x^2 - 7x + 4 = 2 x^2 - 7x + 2 = 0 p + q = -b/a = 7 (Sum of the roots) pq = c/a = 2 (Product of the roots) Recall (p + q)^2 = p^2 + q^2 + 2pq So p^2 + q^2 = 7^2 - 2*2 = 45 |p - q|^2 = p^2 + q^2 - 2pq |p - q|^2 = 45 - 2*2 = 41 |p - q| = \sqrt{41} Answer (E) Quadratic Equations: https://youtu.be/QOSVZ7JLuH0

VivekSri · Answer

OA is E

Posted from my mobile device

KD84 · Answer

You can also get there with zero memorization, just some algebra. Starting with the original equation  x^2-7x+4=2  with  p  and  q :

p^2-7p=-2
q^2-7q=-2

subtract line 2 from line 1:

p^2-q^2 - 7(p-q) = 0
(p+q)(p-q)=7(p-q)   
 (p+q) = 7   
since  |p-q| 
eq{0}

now add those first two equations:

p^2+q^2-7(p+q)=-4
 
subtract  -2pq  from both sides, substitute  7  for  (p+q) :

p^2 - 2pq + q^2 - 7*7 = -4 - 2pq
(p-q)^2 = 45 - 2pq
 
to find   pq , square  (p+q)  and  (p-q)  and subtract:

p^2+2pq+q^2 = 49
p^2-2pq+q^2 = (p-q)^2
4pq = 49-(p-q)^2
 
recall that:
 (p-q)^2 = 45 - 2pq
2(p-q)^2 = 90 - 4pq
2(p-q)^2 = 90 - 49+(p-q)^2
(p-q)^2 = 41 
finally gives us:
 |p-q| = \sqrt{41}

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If the 2 real number solutions of x^2 - 7x + 4 are represented by p

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