Question

Roots of the equation
x²+13x+36=0 are
​

Answer 1

Answer:

Use the quadratic formula

=

−

±

2

−

4

√

2

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

x=2a−b±b2−4ac

Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

2

+

1

3

+

3

6

=

0

x^{2}+13x+36=0

x2+13x+36=0

=

1

a={\color{#c92786}{1}}

a=1

=

1

3

b={\color{#e8710a}{13}}

b=13

=

3

6

c={\color{#129eaf}{36}}

c=36

=

−

1

3

±

1

3

2

−

4

⋅

1

⋅

3

6

√

2

⋅

1

x=\frac{-{\color{#e8710a}{13}} \pm \sqrt{{\color{#e8710a}{13}}^{2}-4 \cdot {\color{#c92786}{1}} \cdot {\color{#129eaf}{36}}}}{2 \cdot {\color{#c92786}{1}}}

x=2⋅1−13±132−4⋅1⋅36

2

Answer 2

Answer:

x=-4 and x=-9

Step-by-step explanation:

x²+ 13x+36

=x²+9x+4x+36

=x(x+9)+4(x+9)

=(x+4)(x+9)

=x+4=0 and x+9=0

=x=-4 and x=-9