Question

The roots of quadratic equation x 2 + 14x + 40 = 0 are:

Answer 1

Step-by-step explanation:

Factoring x2-14x+40

The first term is, x2 its coefficient is 1 .

The middle term is, -14x its coefficient is -14 .

The last term, "the constant", is +40

Step-1 : Multiply the coefficient of the first term by the constant 1 • 40 = 40

Step-2 : Find two factors of 40 whose sum equals the coefficient of the middle term, which is -14 .

-40 + -1 = -41

-20 + -2 = -22

-10 + -4 = -14 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and -4

x2 - 10x - 4x - 40

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-10)

Add up the last 2 terms, pulling out common factors :

4 • (x-10)

Step-5 : Add up the four terms of step 4 :

(x-4) • (x-10)

Which is the desired factorization

Equation at the end of step1

:

(x - 4) • (x - 10) = 0

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Answer 2

Concept Introduction:-

It might resemble a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that $x^2 + 14x + 40 = 0$

To Find:-

We have to find that the roots of quadratic equation $x^2 + 14x + 40 = 0$

Solution:-

According to the problem

$x=\frac{-14 \pm \sqrt{14^{2}-4 \times 40}}{2}\\x=\frac{-14 \pm \sqrt{196-4 \times 40}}{2}\\x=\frac{-14 \pm \sqrt{196-160}}{2}\\x=\frac{-14 \pm \sqrt{36}}{2}\\x=\frac{-14 \pm 6}{2}\\x=\frac{-8}{2}\\x=-4$

Now solve the equation $x=\frac{-14 \pm 6}{2}$ when $\pm$ is minus. Subtract $6$ from $-14$.

$x=\frac{-20}{2}$

Divide $-20$ by $2$.

$x=-10$

The equation is now solved.

$\begin{aligned}&x=-4 \\&x=-10\end{aligned}$

Final Answer:-

The roots are $-4$ and $-10.$

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