I. p^{2} +7p + 10 = 0 II. q^{2}- q – 6 = 0

I. p^{2}+7p + 10 = 0,

p^{2}+ 5p + 2p + 10 + 6 = 0; or

p( p+5) + 2( p+5) = 0

(p+2)( p+5) = 0, or p = -2 or = -5,

II. q^{2} - q – 6 = 0;

q^{2} – 3q + 2q – 6 = 0

q(q-3) + 2(q -3) = 0 or, q = -2 or 3

Hence, q ≥ p

Alternate Method:

if signs of quadratic equation is +ve and +ve respectively then the roots of equation will be -ve and -ve.

So, roots of first equation = p = -2, -5

if signs of quadratic equation is -ve and -ve respectively then the roots of equation will be +ve and -ve. (note: -ve sign will come in smaller root)

So, roots of second equation = q = -2, 3

After comparing roots of quadratic eqution we can conclude that q ≥ p.

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^{2}- 37b + 143 = 0 II. 2a^{2}+ 15a - 143 = 0 - I. 8x – 3y = 85 II. 4x – 5y = 67
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^{2}+ 3x - 9 = 0 II. 3y^{2}- y - 10 = 0 - I. 35 y² + 58 y + 24 = 0 II. 21 x² + 37 x + 12 = 0
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^{3}= 1728 II. y^{2}– 15y + 56 = 0 - I. x²= 961 II. y= √961
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^{2}+ 22x + 8 = 0 II. 4y^{2}- y − 3 = 0 - I. 2x² - 15x + 13 = 0 II. 3y² - 6y + 3 = 0
- I. 2x
^{2 }+ 5x + 2 = 0 II. 4y^{2 }= 1

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