I am not sure if my son got the first answers right and he is stuck on the last three. The teacher did not provide good instruct
ions. Can anyone help us out? Thank you.
1 answer:
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Answer:
2 real solutions
Step-by-step explanation:
Remember this messy thing?
The <em>quadratic formula</em>, as it's called, gives us the roots to any quadratic equation in standard form (ax² + bx + c = 0). The information on the <em>type</em> of roots is contained entirely in that bit under the square root symbol (b² - 4ac), called the <em>discriminant</em>. If it's non-negative, we'll have <em>real</em> roots, if it's negative, we'll have <em>complex roots</em>.
For our equation, we have a discrimant of (-3)² - 4(6)(-4) = 9 + 96 = 105, which is non-negative, so we'll have real solutions, and since quadratics are degree 2, we'll have exactly 2 real solutions.
Answer:
CE = 17
Step-by-step explanation:
∵ m∠D = 90
∵ DK ⊥ CE
∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK
In the two Δ KDE and KCD:
∵ m∠KDE = m∠KCD
∵ m∠DKE = m∠CKD
∵ DK is a common side
∴ Δ KDE is similar to ΔKCD
∴
∵ DE : CD = 5 : 3
∴
∴ KD = 5/3 KC
∵ KE = KC + 8
∵
∴
∴
∴
∴
∴ KC = (8 × 9) ÷ 16 = 4.5
∴ KE = 8 + 4.5 = 12.5
∴ CE = 12.5 + 4.5 = 17