Need help with this math question.?
Given a quadratic equation 2x^2+mx-20=0 where m is constant, find the value of m if
1-one of the roots of the equation is 4
2-the sum of roots of the equation is -2
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answers:
khalil say: 1-
substitute x = 4
2x^2+mx-20=0
2(4)² + m(4) - 20 = 0
4m + 12 = 0
m = -3

2-
ax² + bx + c = 0
sum of roots = -b/a
-m/2 = -2
m = 4
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Krishnamurthy say: Given a quadratic equation
2x^2 + mx - 20 = 0 where m is constant.
find the value of m if
1-one of the roots of the equation is 4
2-the sum of roots of the equation is -2
2(x - 2)(x + 5) = 0
2x^2 + 4x - 20 = 0
m = 4
-
la console say: 2x² + mx - 20 = 0

Polynomial like: ax² + bx + c, where:
a = 2
b = m
c = - 20

Δ = b² - 4ac (discriminant)

Δ = m² - 4.(2 * - 20)

Δ = m² + 160 ← whatever the value of m, you can see that: Δ > 0 → 2 distinct solutions

x₁ = (- b - √Δ) / 2a

x₂ = (- b + √Δ) / 2a





One of the roots of the equation is 4. Suppose that it's x₁

x₁ = 4

(- b - √Δ) / 2a = 4

- b - √Δ = 8a → recall: b = m

- m - √Δ = 8a → recall: a = 2

- m - √Δ = 16

√Δ = - m - 16 → you square both sides

(√Δ)² = (- m - 16)²

Δ = m² + 32m + 256 → recall: Δ = m² + 160

m² + 160 = m² + 32m + 256

32m = - 96

→ m = - 3


One of the roots of the equation is 4. Suppose that it's x₂

x₂ = 4

(- b + √Δ) / 2a = 4

- b + √Δ = 8a → recall: b = m

- m + √Δ = 8a → recall: a = 2

- m + √Δ = 16

√Δ = m + 16 → you square both sides

(√Δ)² = (m + 16)²

Δ = m² + 32m + 256 → recall: Δ = m² + 160

m² + 160 = m² + 32m + 256

32m = - 96

→ m = - 3



The sum of roots of the equation is:

= x₁ + x₂

= [(- b - √Δ) / 2a] + [(- b + √Δ) / 2a]

= [(- b - √Δ) + (- b + √Δ)] / 2a

= [- b - √Δ - b + √Δ] / 2a

= - 2b / 2a

= - b/a → recall: b = m

= - m/a → recall: a = 2

= - m/2 → given that the sum of the roots is - 2

- m/2 = - 2

→ m = 4
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Pinkgreen say: (1) if x=4 is a root, then
2(4)^2+4m-20=0
=>
m=-3

(2) the sum of roots=-2
=>
-2=-m/2
=>
m=4
-
llaffer say: If you know that one of the roots is 4, substitute 4 for x and solve for m:

2x² + mx - 20 = 0
2(4)² + 4m - 20 = 0
2(16) + 4m - 20 = 0
32 + 4m - 20 = 0
12 + 4m = 0
4m = -12
m = -3

For the sum of two roots, the sum of the roots in a generic quadratic equation (ax² + bx + c = 0) will be -b/a

So in your equation we have:

a = 2
b = m

So if the sum of the roots is -2, we get:

S = -b/a
-2 = -m/2
-4 = -m
m = 4
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