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3 years ago

8

Rewrite the equation by completing the square. x^2−4x+3=0

1 answer:

harina [27]3 years ago

3 0

Answer:

x=1 or x=3

Step-by-step explanation:

Hello, please consider the following.

$x^2-4x+3=0\\\\\text{Step 1 - complete the square}\\\\x^2-4x+3=x^2-2*2*x+3=(x-2)^2-4+3=(x-2)^2-1=0\\\\\text{Step 2 - move the constant to the right side, meaning adding 1 here.}\\\\(x-2)^2=1\\\\\text{Step 3 - take the root}\\\\x-2=\pm1\\\\x=2-1=1 \ \ or \ \ x=2+1=3$

Thank you

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What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where

Setler79 [48]

Answer:

$k=-16,k=-8,k=8,k=16$

Step-by-step explanation:

We are given quadratic equations as

$x^2+kx+15$

and it can be factored as

$=(x+a)(x+b)$

now, we can multiply factor term

$(x+a)(x+b)=x^2+(a+b)x+ab$

now, we can compare

$x^2+(a+b)x+ab=x^2+kx+15$

so, we get

$k=a+b$

$ab=15$

we are given that

'a' and 'b' are integers

so, we can find all possible factors

$15=(-1\times -15),(1\times 15)$

$15=(-3\times -5),(3\times 5)$

so, we can find k

At $(-1\times -15)$ :

$k=a+b$

we can plug values

$k=-1-15$

$k=-16$

At $(1\times 15)$ :

$k=a+b$

we can plug values

$k=1+15$

$k=16$

At $(-3\times -5)$ :

$k=a+b$

we can plug values

$k=-3-5$

$k=-8$

At $(3\times 5)$ :

$k=a+b$

we can plug values

$k=3+5$

$k=8$

So, values of k are

$k=-16,k=-8,k=8,k=16$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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or

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