Question

I need help please (algebra 1)

Answer 1

• Option C (x = 3 or x = -13)

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Explanation :

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`\longrightarrow \sf \qquad {x}^(2) + 10x = 39`

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We can write it as,

`\longrightarrow \sf \qquad {x}^(2) + 10x - 39 = 0`

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We have to find the two numbers a and b such that,

`\longrightarrow \sf \qquad a + b = 10`

`\longrightarrow \sf \qquad a b = 39`

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Obviously, the two numbers are 3 and 13.

`\longrightarrow \sf \qquad {x}^(2) - 3x + 13x - 39 = 0`

`\longrightarrow \sf \qquad {x}(x - 3)+ 13(x - 3) = 0`

`\longrightarrow \sf \qquad ({x}+ 13)(x - 3) = 0`

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Whether, the value of x :

`\longrightarrow \sf \qquad {x}+ 13 = 0`

`\longrightarrow \pmb{\bf \qquad {x} = - 13}`

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Whether, the value of x :

`\longrightarrow \sf \qquad {x} - 3 = 0`

`{\longrightarrow { \pmb{\bf \qquad {x} = 3}}}`

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So,

• Option C (x = 3 or x = -13) is correct.