How many real solutions does a quadratic equation have if its discriminant is negative?

Mathematics

1 answer:

kvasek [131]3 years ago

3 0

Answer:

Step-by-step explanation:

Discriminant is given by

$D = b^{2} -4ac$

If it is negative then in quadratic formula the square root part becomes negative which makes the solution complex.

so there cannot be any real solution if the Discriminant is negative or less than 0 ,

The correct option for the given question is

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Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

garri49 [273]

Answer:

BE = 294.23 ft

PE = 259.62 ft

Step-by-step explanation:

Quadrilateral CPRG is a parallelogram.

CP || GR

In triangles BCG and BPE

$\angle BCG \cong \angle BPE (Alternate \: \angle 's) \\\angle GBC \cong \angle EBP (Vertical \: \angle 's) \\\therefore \triangle BCG \sim \triangle BPE (AA \: postulate) \\\\\therefore \frac{BE}{BG} =\frac{PE}{GC} =\frac{BP}{BC}...(csst)\\\\\therefore \frac{BE}{425} =\frac{PE}{375} =\frac{225}{325}\\\\\therefore \frac{BE}{425} =\frac{PE}{375} =\frac{9}{13}\\\\\therefore \frac{BE}{425} =\frac{9}{13}\\\\BE =\frac{9\times 425}{13}\\\\BE =\frac{3,825}{13}\\\\BE = 294.23\: ft \\\\\&\\\frac{PE}{375} =\frac{9}{13}\\\\PE =\frac{9\times 325}{13}\\\\PE =\frac{3,375}{13}\\\\PE = 259.615385 \\\\PE = 259.62 \: ft\\\\$