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ziro4ka [17]
3 years ago
7

Find the value of c that completes the square for x2−15x+c

Mathematics
1 answer:
meriva3 years ago
5 0

Answer:

The value of <em>c</em> is \frac{225}{4}.

Step-by-step explanation:

The perfect square of the difference between two numbers is:

(x-y)^{2}=x^{2}-2xy+y^{2}

The expression provided is:

x^{2}-15x+c

The expression is a perfect square of the difference between two numbers.

One of the number is <em>x</em> and the other is √<em>c</em>.

Use the above relation to compute the value of <em>c</em> as follows:

x^{2}-15x+c=(x-\sqrt{c})^{2}\\\\x^{2}-15x+c=x^{2}-2\cdot x\cdot\sqrt{c}+c\\\\15x=2\cdot x\cdot\sqrt{c}\\\\15=2\cdot\sqrt{c}\\\\\sqrt{c}=\frac{15}{2}\\\\c=[\frac{15}{2}]^{2}\\\\c=\frac{225}{4}

Thus, the value of <em>c</em> is \frac{225}{4}.

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X-y=5 and x^2y=5x+6​
sergeinik [125]

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

<h3>How to solve a system of equations</h3>

In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:

x - y = 5      (1)

x² · y = 5 · x + 6       (2)

By (1):

y = x + 5

By substituting on (2):

x² · (x + 5) = 5 · x + 6

x³ + 5 · x² - 5 · x - 6 = 0

(x + 5.693) · (x - 1.430) · (x + 0.737) = 0

There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737

And the y-values are found by evaluating on (1):

y = x + 5

x₁ ≈ 5.693

y₁ ≈ 10.693

x₂ ≈ 1.430

y₂ ≈ 6.430

x₃ ≈ - 0.737

y₃ ≈ 4.263

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

To learn more on nonlinear equations: brainly.com/question/20242917

#SPJ1

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You take 1.75, and subtract 1.48 from it.     1.75

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