Question

m²-m-1=0 find the quadratic equations​

Answer 1

Step-by-step explanation:

m^2-m-1=0

a=1 b=-1 c=-1

$m = - b \binom{ + }{ - } \sqrt{b { }^{2} } - 4ac \div 2a$

1(+-)

$1 \binom{ + }{ - } \sqrt{( - 1) {}^{2} } - 4(1)( - 1) \div 2(1) \\ 1 \binom{ + }{ - } \sqrt{1 + 4 } \div 2 \\ \\ 1 \binom{ + }{ - } \sqrt{5 } \div 2 \\ case1 = 1 + \sqrt{5 } \div 2 \\ case2 = 1 - \sqrt{5} \div 2$

Answer 2

Answer:

$\Large\boxed{M=\frac{1+\sqrt{5}}{2},\:M=\frac{1-\sqrt{5}}{2}}$

Step-by-step explanation:

GIVEN:

$\Large\boxed{\textnormal{LESSON: QUADRATIC EQUATIONS}}$

$\Large\boxed{\textnormal{SUBJECT: MATH }}}$

Isolate by the m from one side of the equation.

SOLUTIONS:

First, you have to use quadratic equations formula.

$\Large\boxed{\textnormal{QUADRATIC EQUATIONS FORMULA}}$

$\displaystyle AX^2+BX+C=0$

$\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2A}$

A=1

B=(-1)

C=(-1)

Rewrite the problem down.

$\displaystyle \frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4*\:1\left(-1\right)}}{2*\:1}$

Solve. (Simplify to find the answer!)

$\displaystyle \frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4*\:1\left(-1\right)}}{2*\:1}=\frac{1+\sqrt{5} }{2}=\frac{1-\sqrt{5} }{2}$

$\displaystyle \boxed{\frac{1+\sqrt{5} }{2}=\frac{1-\sqrt{5} }{2} }}$

$\Large\boxed{M=\frac{1+\sqrt{5} }{2},\quad M=\frac{1-\sqrt{5} }{2} }$

Therefore, the final answer is m=1+√5/2, m=1-√5/2.