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If integers p and q are the roots of the equation ax^2 + bx + c = 0, where a, b and c are constants and a > 0, by what percentage is c greater than |b|?
(1) |p+1| = |q – 3|
(2) The greatest number that divides both |p| and |q| is 2 and the smallest number that is divisible by both |p| and |q| is 12
(1) |p+1| = |q – 3|
(2) The greatest number that divides both |p| and |q| is 2 and the smallest number that is divisible by both |p| and |q| is 12
20 Jun 2017, 11:51
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We have quadratic equation ax^2 + bx + c = 0(where a, b and c are constants and a>0)
We need to find out what percentage is 'c' greater than |b|(absolute value of b). Also known are the roots of the roots are p and q.
(1) |p+1| = |q–3|
Assume values for p and q which satisfy the equation.
Example 1 : p=+2,q=+5. A quadratic equation can be formed by (x-p)(x-q)
(x-2)(x-5) => x^2 -7x +10 = 0. Here c(10) is around 42.85% greater than |b|, which is 7.
Example 2 : p=+2,q=0. A quadratic equation can be formed by (x-p)(x-q)
(x-2)(x) => x^2 -2x +1 = 0 Here c(1) is around 50% lesser than |b|, which is 2. Clearly insufficient.
(2) The greatest number that divides both |p| and |q| is 2 and the smallest number that is divisible by both |p| and |q| is 12
Example 1 : p=-2,q=-6 A quadratic equation can be formed by (x-p)(x-q)
(x+2)(x+6) => x^2+4x+12 Here c(12) is around 200% greater than |b|, which is 4.
Example 2: p=-4,q=-6 A quadratic equation can be formed by (x-p)(x-q)
(x+4)(x+6) => x^2+10x+24 Here c(24) is around 140% greater than |b|, which is 10.Clearly insufficient.
When we combine both the statements, The only value possible for p & q, is p=-4,q=6
When p=-4,q=6 the conditions in both the statements are satisfied and the quadratic equation x^2 - 2x - 24=0
This has value of c(24) which is 1100% greater than |b|. Hence the combination of these statements is sufficient(Option C)
We need to find out what percentage is 'c' greater than |b|(absolute value of b). Also known are the roots of the roots are p and q.
(1) |p+1| = |q–3|
Assume values for p and q which satisfy the equation.
Example 1 : p=+2,q=+5. A quadratic equation can be formed by (x-p)(x-q)
(x-2)(x-5) => x^2 -7x +10 = 0. Here c(10) is around 42.85% greater than |b|, which is 7.
Example 2 : p=+2,q=0. A quadratic equation can be formed by (x-p)(x-q)
(x-2)(x) => x^2 -2x +1 = 0 Here c(1) is around 50% lesser than |b|, which is 2. Clearly insufficient.
(2) The greatest number that divides both |p| and |q| is 2 and the smallest number that is divisible by both |p| and |q| is 12
Example 1 : p=-2,q=-6 A quadratic equation can be formed by (x-p)(x-q)
(x+2)(x+6) => x^2+4x+12 Here c(12) is around 200% greater than |b|, which is 4.
Example 2: p=-4,q=-6 A quadratic equation can be formed by (x-p)(x-q)
(x+4)(x+6) => x^2+10x+24 Here c(24) is around 140% greater than |b|, which is 10.Clearly insufficient.
When we combine both the statements, The only value possible for p & q, is p=-4,q=6
When p=-4,q=6 the conditions in both the statements are satisfied and the quadratic equation x^2 - 2x - 24=0
This has value of c(24) which is 1100% greater than |b|. Hence the combination of these statements is sufficient(Option C)
General Discussion
