Solve this with your work shown. (50 POINTS.)
1 answer:
top problems:
1-) x²-2x-8
<h2>(x-4)(x+2)</h2>---------------------
2-) 4s²-4s+1
<h2>(2s-1)²</h2>---------------------
3-) 6x²-3-7x
<h2>(2x-3)(3x+1)</h2>---------------------
4-) 4u²+8u
<h2>4u(u+2)</h2>---------------------
5-) t²-15
<h2>2t</h2>---------------------
6-) 3x²-x-4
<h2>(3x-4)(x+1)</h2>bottom problems:
1-)
y=x²+
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2-)
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3-) 0, √5
<h2>y=x²-x√5</h2>---------------------
4-) 1,1
<h2>y= x²-2x+1</h2>---------------------
5-)
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6-) 4,1
<h2>y=x²-5x+4</h2>You might be interested in
Answer:
In a quadratic equation of the shape:
y = a*x^2 + b*x + c
we hate that the discriminant is equal to:
D = b^2 - 4*a*c
This thing appears in the Bhaskara's formula for the roots of the quadratic equation:
You can see that the determinant is inside a square root, this means that if D is smaller than zero we will have imaginary roots (the graph never touches the x-axis)
If D = 0, the square root term dissapear, and this implies that both roots of the equation are the same, this means that the graph touches the x axis in only one point, wich coincides with the minimum/maximum of the graph)
If D > 0 we have two different roots, so the graph touches the x-axis in two different points.