Question

a²-11a-42 please give me answer​

Answer 1

Step-by-step explanation:

So,

a²-11a-42

p + q = -42

pq = 1 (-42) = -42

1, -42

2, -21

3, -14

4, -7

Calculate the sum for each pair.

1−42=−41

2−21=−19

3−14=−11

6−7=−1

The solution is the pair that gives sum −11.

p=−14

q=3

Rewrite $a^2$−11a−42 as ($a^2-14a$)+(3a−42).

($a^2-14a$) + (3a−42)

a(a−14) + 3(a−14)

So here we came with the conclusion that:

(a-14) (a+3) is your required answer

Answer 2

Given: Quadratic Equation $a^{2} - 11a-42$

To Find: The roots/zeroes of the Quadratic Equation $a^{2} - 11a-42$.

Solution:

Quadratic Equation $a^{2} - 11a-42$

• Here, we will solve this question by using the quadratic formula $\frac{-b\pm\sqrt{b^{2} - 4ac} }{2a}$  (Note - You can solve this by splitting the middle term too.)

• Identify the coefficients a,b and c of the quadratic equation.

$a=1, b=-11, c=-42$

• Substitute $a=1, b=-11, c=-42$ into the quadratic formula $\frac{-b\pm\sqrt{b^{2} - 4ac} }{2a}$.

$a=\frac{-(-11)\pm\sqrt{(-11)^{2}-4\times1\times(-42) } }{2\times1}$

• Any expression multiplied by $1$ remains the same:

$a=\frac{-(-11)\pm\sqrt{(-11)^{2}-4\times(-42) } }{2\times1}$

• Any expression multiplied by $1$ remains the same:

$a=\frac{-(-11)\pm\sqrt{(-11)^{2}-4\times(-42) } }{2}$

• When two minus are close, they turn into plus.

$a=\frac{11\pm\sqrt{(-11)^{2}-4\times(-42) } }{2}$

• Evaluate the power.

$a=\frac{11\pm\sqrt{(121)-4\times(-42) } }{2}$

• Multiply the numbers

$a=\frac{11\pm\sqrt{121+168 } }{2}$

• Add the numbers:

$a=\frac{11\pm\sqrt{289 } }{2}$

• Evaluate the square root:

$a=\frac{11\pm17 }{2}$

• Separate the solutions:

$a = \frac{11+17}{2} \\a = \frac{11-17}{2}$

• Solve for $a$

$a = \frac{28}{2} \\a = \frac{-6}{2}$

• Divide now:

$a = 14\\a = -3$

Final Answer:

Hence, the roots of the quadratic equation $a^{2} - 11a-42$ are $14$ and $-3$.