Show that the equation x^3 +6x-5=0 has a solution between x=0 and x= 1
1 answer:
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<h2>Step-by-step explanation:</h2>
Given equations;
y₁ = 3x - 8 -------------------(i)
y₂ = 0.5x + 7 --------------------(ii)
To fill the table, substitute the values of x into equations (i) and (ii)
=> At x = 0
y₁ = 3(0) - 8 = -8
y₂ = 0.5(0) + 7 = 7
=> At x = 1
y₁ = 3(1) - 8 = -5
y₂ = 0.5(1) + 7 = 7.5
=> At x = 2
y₁ = 3(2) - 8 = -2
y₂ = 0.5(2) + 7 = 8
=> At x = 3
y₁ = 3(3) - 8 = 1
y₂ = 0.5(3) + 7 = 8.5
=> At x = 4
y₁ = 3(4) - 8 = 4
y₂ = 0.5(4) + 7 = 9
=> At x = 5
y₁ = 3(5) - 8 = 7
y₂ = 0.5(5) + 7 = 9.5
=> At x = 6
y₁ = 3(6) - 8 = 10
y₂ = 0.5(6) + 7 = 10
=> At x = 7
y₁ = 3(7) - 8 = 13
y₂ = 0.5(7) + 7 = 10.5
=> At x = 8
y₁ = 3(8) - 8 = 16
y₂ = 0.5(8) + 7 = 11
=> At x = 9
y₁ = 3(9) - 8 = 19
y₂ = 0.5(9) + 7 = 11.5
=> At x = 10
y₁ = 3(10) - 8 = 22
y₂ = 0.5(10) + 7 = 12
The complete table is attached to this response.
(ii) To find the solution of the system of equations using the table, we find the value of x for which y₁ and y₂ are the same.
As shown in the table, that value of <em>x = 6</em>. At this value of x, the values of y₁ and y₂ are both 10.