Question

solve this quadratic equations

Answer 1

x^2-(16-5)x-80
x^2-16x+5x-80
X(x-16)+5(x-16)
(x-16) (X+5)

2.
x^2-4x-45
x^2-(9-5)x-45
x^2-9x+5x-45
X(x-9)+5(x-9)
(x-9) (X+5)

3.
x^2+13x+42
x^2+(7+6)X+42
x^2+7x+6x+42
X(X+7)+6(X+7)
(X+7) (X+6)

4.
x^2-16x+39
x^2-(13+3)X+39
x^2-13x-3x+39
X(x-13)-3(x-13)
(x-13)(x-3)

hope it help you

Answer 2

Question

1)x²-11x-80=0

2)x²-4x-45=0

3)x²+13x+42=0

4)x²-16x+39=0

To find

factorise the Equation

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

We use quadratic formula for finding the roots :

1)x²-11x-80=0

a= 1, b= -11 & c= -80

x= -b±√b²-4ac /2a

=>x= -(-11)±√(-11)²-4(1)(-80) / 2(1)

=>x= 11±√121-(-320) /2

=>x= 11±√441 / 2

=>x= The discriminant b²-4ac is > 0

so,there are two real roots.

=>x=11±21 /2

=>x= 32/2 or x= -10/2

=>x= 16 or x= -5

x= 16 and x = -5

2) x²-4x-45=0

a= 1 ,b = -4 & c= -45

=>x= -(-4)±√(-4)²-4(1)(-45) / 2

=>x= 4±√16-(-180) / 2

=>x= 4±√196/2

Here,the discriminant b²-4ac is >0

So,there are two real roots.

=>x= 4±14/2

=>x= 18/2 or -10/2

x = 9 or x= -5

x= 9 and x= -5

3)x²+13x+42=0

a= 1 ,b = 13 ,c= 42

=>x= -13±√(13)²-4(1)(42) / 2

=>x = -13±√169-168 /2

=>x= -13±√1 /2

Here,the discriminant b²-4ac is >0

So,there are two real roots.

=>x= -13±1/2

=>x= -12/2 or x = -14/2

x= -6 or x= -7

4)x²-16x+39=0

a= 1 ,b = -16 & c=39

=>x= -(-16)±√ (-16)²-4(1)(39) / 2

=>x= 16±√256-156/2

=>x=16±√100/2

Here,the discriminant b²-4ac is >0

So,there are two real roots.

=>x= 16±10/2

x = 26/2 or x= 6/2

x=13 or x= 3