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kotykmax [81]
3 years ago
5

Solve each quadratic equation by factoring and check

Mathematics
1 answer:
max2010maxim [7]3 years ago
5 0

hello : <span>
<span>the discriminat of each quadratic equation : ax²+bx+c=0 ....(a ≠ 0) is :
Δ = b² -4ac
1 )  Δ > 0  the equation has two reals solutions : x1,2 =  (-b±√Δ)/2a</span></span>

factoring :

<span>ax²+bx+c = a(x-x1)(x-x2)
2 ) Δ = 0 : one solution : x1 = x2 = -b/2a</span>

factoring  : 

<span>ax²+bx+c = a(x-x1)²
3 ) Δ < 0 : no reals solutions... no </span>

<span>factoring </span>
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3 years ago
Based on the information given say whether or not △ABC∼△FED. Explain your reasoning.
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Answer:

Yes, △ABC ∼ △FED by AA postulate.

Step-by-step explanation:

Given:

Two triangles ABC and FED.

m∠A = m∠B

m∠C = m∠A + 30°

m∠E = m∠F = x

m∠D = 2x-20°.

Now, let m∠A = m∠B = y

So, m∠C = m∠A + 30° = y+30

Now, sum of all interior angles of a triangle is 180°. Therefore,

m∠A + m∠B +  m∠C = 180

y+y+y+30=180\\3y=180-30\\3y=150\\y=\frac{150}{3}=50

Therefore, m∠A = 50°, m∠B = 50° and m∠C =  m∠A + 30° = 50 + 30 = 80°.

Now, consider triangle FED,

m∠D+ m∠E + m∠F = 180

2x-20+x+x=180\\4x=180+20\\4x=200\\x=\frac{200}{4}=50

Therefore,  m∠F = 50°  

m∠E = 50° and  

m∠D =  2x-20=2(50)-20=100-20=80\°

So, both the triangles have congruent corresponding angle measures.

m∠A = m∠F = 50°

m∠B = m∠E = 50°

m∠C = m∠D = 80°

Therefore, the two triangles are similar by AA postulate.

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Answer:

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Step-by-step explanation:

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