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3 years ago

4 times a number is 21 less than the square of that number. Find the positive solution

1 answer:

8 0

set up the equation 4x + 21 = x^2. getting everything on one side so you have x^2-4x-21= 0. find two numbers that multiply to -21 and add to -4, which are 3 and -7. substitute in with them as x to get x^2+3x-7x+21 = 0. factor out the x and the 7 to get x(x+3)-7(x+3) = 21. so x = -3 and 7. the positive solution would be 7.

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3 years ago

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Softa [21]

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3 years ago

Determine the type and number of solutions of x^2 + 10x − 25 = 0.

Vitek1552 [10]

Answer:

There is exactly 1 solution and it is a double root.

Step-by-step explanation:

To find this, factor the equation and solve.

x^2 + 10x - 25 = 0

(x - 5)(x - 5) = 0

Now that we have this factored, we can set each parenthesis equal to 0 and solve separately.

x - 5 = 0

x = 5

x - 5 = 0

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Because there are two of the exact same answer, we know it to be a double root.

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4 years ago

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Mekhanik [1.2K]

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3 years ago

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zysi [14]

Answer:

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Step-by-step explanation:

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3 years ago

4 times a number is 21 less than the square of that number. Find the positive solution ​

4 times a number is 21 less than the square of that number. Find the positive solution