Question

i need help please (algebra 1)

Answer 1

Answer :

• Option C (x = 3 or x = -13)

⠀

Explanation :

⠀

$\longrightarrow \sf \qquad {x}^{2} + 10x = 39$

⠀

We can write it as,

$\longrightarrow \sf \qquad {x}^{2} + 10x - 39 = 0$

⠀

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

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

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

⠀

Obviously, the two numbers are 3 and 13.

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

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

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

⠀

Whether, the value of x :

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

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

⠀

Whether, the value of x :

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

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

⠀

So,

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