Question

Разложите на множители квадратный трехчлен : а) х^-15x+26 б)4y^+3y-7

Answer 1

$a)\\\\x^{2}-15x+26=0\\\\x_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{15-\sqrt{15^{2}-4*1*26}}{2*1}=\frac{15-\sqrt{121}}{2}=\frac{15-11}{2}=2\\\\x_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}=\frac{15+\sqrt{15^{2}-4*1*26}}{2*1}=\frac{15+\sqrt{121}}{2}=\frac{15+11}{2}=13\\\\ax^{2}+bx+c=a*(x-x_{1})(x-x_{2})\\\\x^{2}-15x+26=(x-2)(x-13)\\\\b)\\\\4y^{2}+3y-7=0\\\\x_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{-3-\sqrt{3^{2}-4*4*(-7)}}{2*4}=\frac{-3-\sqrt{121}}{8}=\frac{-3-11}{8}=-\frac{7}{4}\\\\$$x_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}=\frac{-3+\sqrt{3^{2}-4*4*(-7)}}{2*4}=\frac{-3+\sqrt{121}}{8}=\frac{-3+11}{8}=1\\\\ax^{2}+bx+c=a*(x-x_{1})(x-x_{2})\\\\4y^{2}+3y-7=4*(x+\frac{7}{4})(x-1)=(4x+7)(x-1)\\\\$