Mathematics, 05.10.2019 07:10 fantasticjrod

What substitution should be used to rewrite 6(x + 5)2 + 5(x + 5) – 4 = 0 as a quadratic equation? u = (x + 5) u = (x – 5) u = (x + 5)2 u = (x – 5)2

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Answers

The quadratic equation $6(x+5)^{2}+5(x+5)-4=0$ can be rewritten by substituting $\boxed{u=(x+5)}$ i.e., option (1) is correct.

Further explanation:

The equation of the form $ax^{2}+bx+c=0$ is defined as quadratic equation where a,b and are real numbers.

The given quadratic equation is as follows:

$\boxed{6(x+5)^{2}+5(x+5)-4=0}$

Option (1)

Here, the option is u=(x+5) .

Substitute u=(x+5) in the equation $6(x+5)^{2}+5(x+5)-4=0$ .

$\begin{aligned}6(u)^{2}+5(u)-4&=0\\6u^{2}+5u-4&=0\end{aligned}$

The above obtained equation is the quadratic equation.

Therefore, option (1) is correct.

Option (2)

Here, the option is u=x-5 or u+5=x .

Substitute u+5=x in the equation $6(x+5)^{2}+5(x+5)-4=0$ .

$\begin{aligned}6(u+5+5)^{2}+5(u+5+5)-4&=0\\6(u+10)^{2}+5(u+10)-4&=0\end{aligned}$

The above obtained equation is a quadratic equation but the given equation $6(x+5)^{2}+5(x+5)-4=0$ cannot be rewritten as $6(u+10)^{2}+5(u+10)-4=0$ .

Therefore, option (2) is incorrect.

Option (3)

Here, the option is $u=(x+5)^{2}$ .

Substitute $u=(x+5)^{2}$ in the given equation $6(x+5)^{2}+5(x+5)-4=0$ .

$6u+5\sqrt{u}-4=0$

The above obtained equation is not a quadratic equation.

Therefore, option (3) is incorrect.

Option (4)

Here, the option is $u=(x-5)^{2}$ or $\sqrt{u}+5=x$ .

Substitute $\sqrt{u}+5=x$ in the given equation $6(x+5)^{2}+5(x+5)-4=0$ .

$\begin{aligned}6(\sqrt{u}+5+5)^{2}+5(\sqrt{u}+5+5)-4&=0\\6(\sqrt{u}+10)^{2}+5(\sqrt{u}+10)-4&=0\end{aligned}$

The above obtained equation is the not in the form of $ax^{2}+bx+c=0$ . Thus, the above equation is not a quadratic equation.

Therefore, option (4) is incorrect.

Thus, the quadratic equation $6(x+5)^{2}+5(x+5)-4=0$ can be rewritten by substituting $\boxed{u=(x+5)}$ i.e., option (1) is correct.

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Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Quadratic equations

Keywords: Quadratic equation, substitution, roots of the equation, function, sets, real numbers, integers, ordinates, abscissa, interval, open interval, closed intervals, semi-closed intervals, semi-open, interval.

Answer from: Quest

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)

Answer from: Quest

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

$f(x) =\sqrt{x}$

this can also be written as

$y^{2} =x$

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

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