The discriminant

Algebra Level 1

The difference of the roots of the quadratic equation x 2 + b x + c = 0 x^2 + bx + c = 0 is -2. Find the discriminant of the given equation.


The answer is 4.

Sakanksha Deo by Sakanksha Deo , 23, India

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5 solutions

Rama Devi
May 13, 2015

14 Helpful 0 Interesting 0 Brilliant 0 Confused

Did the same way!😀

Anurag Pandey - 4 years, 10 months ago
Caleb Townsend
Mar 21, 2015

r = − b ± b 2 − 4 c 2 r 2 − r 1 = − b + b 2 − 4 c 2 − − b − b 2 − 4 c 2 = b 2 − 4 c r 2 − r 1 = 2 b 2 − 4 c = 2 2 = 4 r = \frac{-b \pm \sqrt{b^2 - 4c}}{2} \\ r_2 - r_1 =\frac{-b + \sqrt{b^2 - 4c}}{2}-\frac{-b - \sqrt{b^2 - 4c}}{2} \\ = \sqrt{b^2 - 4c} \\ r_2 - r_1 = 2 \\ b^2 - 4c = 2^2 = \boxed{4}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

For Vieta's formulas, if we call the roots,a and a+2, then c = a ⋅ ( a + 2 ) = a 2 + 2 a c = a \cdot (a +2) = a^2 + 2a and b = − ( 2 a + 2 ) ⇒ b 2 − 4 c = 4 a 2 + 4 + 8 a − 4 a 2 − 8 a = 4 b = - ( 2a +2 ) \Rightarrow b^2 - 4c = 4a^2 + 4 + 8a - 4a^2 - 8a = 4

1 Helpful 0 Interesting 0 Brilliant 1 Confused

Difference of roots = − D a = − 2 = \frac {-\sqrt D}{a} = -2

⟹ − D 1 = − 2 \implies \frac{-\sqrt D}{1} = -2

⟹ D = 2 \implies \sqrt D = 2

∴ D = 4 \therefore D = 4

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Vyom Jain
Apr 16, 2015

difference of roots is equal to (root D)/a (where a is coeff. of x^2)

which gives D equal to 4

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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