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Alisiya [41]
3 years ago
10

Find all real numbers z for which the equation (z-5)x^2-zx+5=0 only has one real solution. If you find more than one, then list

the values separated by commas.
Mathematics
1 answer:
Nataliya [291]3 years ago
7 0

Step-by-step explanation:

For the quadratic equation to have 1 repeated real solution, the discriminant b² - 4ac must be zero.

=> (-z)² - 4(z - 5)(5) = 0

=> z² - 20(z - 5) = 0

=> z² - 20z + 100 = 0

=> (z - 10)² = 0

Therefore z = 10.

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<h3>Answer:  x = (y-2)^2 + 5</h3>

In other words, y-2 goes in the first box and 5 goes in the second box.

=================================================

Work Shown:

y^2 - 4y - x + 9 = 0

y^2 - 4y + 9  = x

x = y^2 - 4y + 9

x = y^2 - 4y + 4 + 5 .... rewrite 9 as 4+5

x = (y^2-4y+4) + 5

x = (y-2)^2 + 5  .... apply the perfect square factoring rule

So we'll have y-2 go in the first box and 5 goes in the second box

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