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secondary 3 | E Maths
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Date Posted: 1 year ago
Views: 338
(a)
x₁ = -2.866
x₂ = 1.386
x₁ = -2.866
x₂ = 1.386
See 2 Answers
Part 1
Imagine LHS is one graph, RHS is one graph.
So you always want to rewrite the equation such that LHS is the graph you have.
2x^2 + 3x - 8 = 0
So Y = 2x^2 + 3x - 8 is the LHS.
The drawn graph is Y = -2x^2 - 3x + 8, which is the negative version of the equation they are asking for, i.e. y = - (2×^2 + 3x - 8)
So you can imagine that u multiply both sides by -1.
Y = 0 is the RHS.
Multiplied by -1 is still 0.
Y= 0 drawn as a line is exactly the x-axis.
So we are looking for when the graph intersects the x-axis. That is the answer. (X = -2.9 or 1.4)
Part 2
Same thing, manipulate this equation so that left hand side = the graph we have.
-2x^2 - 3x + 5 = 0
Add 3 to both sides.
- 2x^2 - 3x + 8 = 3
So we need to draw a line that is y = 3.
And the answer is where this line intersects our graph. (X = -2.5 or 1)
Imagine LHS is one graph, RHS is one graph.
So you always want to rewrite the equation such that LHS is the graph you have.
2x^2 + 3x - 8 = 0
So Y = 2x^2 + 3x - 8 is the LHS.
The drawn graph is Y = -2x^2 - 3x + 8, which is the negative version of the equation they are asking for, i.e. y = - (2×^2 + 3x - 8)
So you can imagine that u multiply both sides by -1.
Y = 0 is the RHS.
Multiplied by -1 is still 0.
Y= 0 drawn as a line is exactly the x-axis.
So we are looking for when the graph intersects the x-axis. That is the answer. (X = -2.9 or 1.4)
Part 2
Same thing, manipulate this equation so that left hand side = the graph we have.
-2x^2 - 3x + 5 = 0
Add 3 to both sides.
- 2x^2 - 3x + 8 = 3
So we need to draw a line that is y = 3.
And the answer is where this line intersects our graph. (X = -2.5 or 1)
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