Question

4x^2+x-33=0, y^2-10y-24=0, p^2+p-90=0

Answer 1

$1)\; \; 4x^2+x-33=0\\\\x^2+\frac{1}{4}x-\frac{33}{4}=0\\\\\Big (x^2+2\cdot \frac{1}{8}x+(\frac{1}{8})^2\Big )-(\frac{1}{8})^2-\frac{33}{4}=0\\\\ (x+\frac{1}{8})^2=\frac{529}{64}\; \; \; \Rightarrow \; \; \; x+\frac{1}{8}=\pm \frac{23}{8}\\\\a)\; \; x+\frac{1}{8}=-\frac{23}{8}\; ,\; \; x=-3\\\\b)\; \; x+\frac{1}{8}=\frac{23}{8}\; ,\; \; x=\frac{22}{8}=\frac{11}{4}$

$2)\; \; y^2-10y-24=0\\\\(y^2-2\cdot 5y+5^2)-5^2-24=0\\\\(y-5)^2=49\; \; \; \Rightarrow \; \; \; y-5=\pm 7\\\\a)\; \; y-5=-7\; ,\; \; x\y=-2\\\\b)\; \; y-5=7\; ,\; \; y=12$

$3)\; \; p^2+p-90=0\\\\\Big (p^2+2\cdot \frac{1}{2}p+(\frac{1}{2})^2\Big )-(\frac{1}{2})^2-90=0\\\\ (p+\frac{1}{2})^2=\frac{361}{4}\\\\a)\; \; p+\frac{1}{2}=-\frac{19}{2}\; ,\; \; p=-\frac{20}{2}=-10\\\\b)\; \; p+\frac{1}{2}=\frac{19}{2}\; ,\; \; p=\frac{18}{2}=9\\\\\\\star \; \; x^2\pm px+q=(x\pm \frac{p}{2})^2-(\frac{p}{2})^2+q\; \; \star$

Answer 2

4x²+x-33=0

D = 1 + 528 = 529

x₁ = (-1 + 23)/8 = 22/8 = 2,75

x₂ = (-1-23)/8 = -3

ответ: -3; 2,75

y²-10y-24=0

х₁ = 12

х₂ = -2, по теореме Виета

ответ: -2; 12

p²+p-90=0

D = 1 + 360 = 361

x₁ = (-1 + 19)/2 = 9

x₂ = (-1-19)/2 = -10

ответ: -10;9