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2 years ago

What two square roots does 40 squared lies between?

Mathematics

2 answers:

masha68 [24]2 years ago

8 0

Answer:

6 and 7

Step-by-step explanation:

√36=6

√49=7

40 is between 36 and 49. Therefore 40 squared lies between the square roots of 6 and 7.

skelet666 [1.2K]2 years ago

8 0

The answers here is 6 and 7

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Find the range of p in equation p(x+1) (x-3)=x-4p-2 has no real roots.

Tanya [424]

Answer:

$\displaystyle p > \frac{1}{4}$ .

Step-by-step explanation:

Expand the left-hand side of this equation:

$p\, (x + 1)\, (x - 3) = p\, \left(x^2 - 2\, x - 3\right) = p\, x^2 - 2\, p\, x - 3\, p$ .

Collect the terms, so that this quadratic equation is in the form $a\, x^2 + b\, x+ c = 0$ :

$p\, x^2 - 2\, p\, x - 3\, p = x - 4\, p - 2$ .

$p\, x^2 - (2\, p + 1)\, x + (p + 2) = 0$ .

In this equation:

$a = p$ .
$b = -(2\, p + 1)$ .
$c = p + 2$ .

Calculate the quadratic discriminant of this quadratic equation:

$\begin{aligned}b^2 - 4\, a\, c &= (-(2\, p + 1))^2 - 4\, p\, (p + 2) \\ &= 4\, p^2 + 4\, p + 1 - 4\, p^2- 8\, p = -4\, p + 1\end{aligned}$ .

A quadratic equation has no real root if its quadratic discriminant is less then zero. As a result, this quadratic equation will have no real root when $-4\, p + 1 < 0$ . Solve for the range of $p$ :

$\displaystyle p > \frac{1}{4}$ .

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2 years ago