X²+11x+q=0
x₁=-7
x₂=? q=?
(-7)²+11*(-7)+q=0
49-77+q=0
-28+q=0
q=28
x²+11x+28=0
D=11²-4*28=121-112=9
x₁=(-11+3)/2=-8/2=-4
x₂=(-11-3)/2=-14/2=-7
Ответ: x₂=-4, q=28
4х*2-25х+36=4(х-4)(х-а)
-17х+36=(4х-8)*(4х-4а)
-17х+36=16х^2-16ха-32х+16а
а=13
<span>1) (4х+3)^-24х=16x^+24x+9-24x=16x^+9</span>
<span>2) <span>(2х-5)^+20х=4x^-20x+25+20x=4x^+25</span></span>
( 2х + 3 )^2 + 7( 2х + 3 ) = 8
2х + 3 = а
а^2 + 7а - 8 = 0
D = 49 + 32 = 81 = 9^2
a1 = ( - 7 + 9 ) : 2 = 1
a2= ( - 7 - 9 ) : 2 = - 8
1) 2x + 3 = 1
2x = - 2
X = - 1
2) 2x + 3 = - 8
2x = - 11
X = - 5,5
Ответ ( - 1 ) ; ( - 5,5 )