How would one find the discriminant of a quadratic equation?

How would one find the discriminant of a quadratic equation?
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Description : Do you know of any website that can solve quadratic inequalities?

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The value of quadratic polynomial f (x) = 2x 2 – 3x- 2 at x = -2 is ...... (a) 12 (b) 15 (c) -12 (d) 16

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Description : If ̳α ‘ and ̳β ‘ are the zeroes of a quadratic polynomial x 2 − 5x + b and α − β = 1, then the value of ̳b‘ is (a) – 5 (b) 6 (c) 5 (d) – 6

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Last Answer : answer:y^x = x^y xln(y) = yln(x) x/ln(x) = y/ln(y) Edit: whoops, I didn’t read thoroughly. You want only one x, not x’s and y’s separated. Sorry about that. It’s not always possible to isolate a variable in an equation, so that could possibly be the case here.

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Last Answer : You’ve only got half the answer I’m afraid. x= -1.618 is also a solution (in fact x= 0.618 and -1.618 at the same time).

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Last Answer : f) 4x - 6y = - 18

The pair of equation x = 4 and y = 3 graphically represents lines which are e) Parallel f) Intersecting at ( 3,4) g) Coincident h) Intersecting at ( 4,3)

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One equation of a pair of dependent linear equation is -2x + 3y = 9 , the second equation can be a) 4x + 6y = 18 b) 4x - 6y = - 18 c) -4x – 6y = 18 d) None of these

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Last Answer : b) 4x - 6y = - 18

The pair of equation x = 4 and y = 3 graphically represents lines which are a) Parallel b) Intersecting at ( 3,4) c) Coincident d) Intersecting at ( 4,3)

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Last Answer : (c) x + 2y =

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Last Answer : (d) 6x + 9y = 21

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Last Answer : (d) -10x+14y-4=0

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In quadratic equation ax^2 + bx + c = 0 , Identities the sum and product of Roots? -Maths 9th

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Last Answer : If the two roots of the quadratic equation ax2 + bx + c = 0 obtained by the quadratic formula be denoted by a and b, then we have α = \(\frac{-b+\sqrt{b^2-4ac}}{2a},\) β = \(\frac{-b-\sqrt{b^2-4ac}}{2a}\) ∴ Sum of roots ... then α + β = - (- \(\frac{5}{6}\)) = \(\frac{5}{6}\), ab = \(\frac{7}{6}\).

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If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

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Last Answer : Given p,q,r are in A.P. then q=2p+r​.....(1). Now px2+qx+r=0 will have real root then q2−4pr≥0. or, 4(p+r)2​−4pr≥0 or, p2+r2−14pr≥0 or, r2−14rp+49p2≥48p2 or, (r−7p)2≥(43​p)2 or, (pr​−7)2≥(43​)2 [ Since p=0 for the given equation to be quadratic] or, ∣∣∣∣∣​pr​−7∣∣∣∣∣​≥43​.

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If the sum as well as the product of roots of a quadratic equation is 9, then the equation is: -Maths 9th

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