First, you need to rewrite the expression into binomial form, so you are working with two terms (as you world with a quadratic):
(x²)²-3(x²)-4=0
Now, you can place the x²s into brackets as the coefficient is now 1:
(x² )(x² )
Next, find out two numbers that add together to give you -3 and multiply to give -4 (these are the leftover integers after removing the x²s). These two numbers are -4 and 1.
Place the -4 and 1 into the brackets:
(x²-4)(x²+1)=0
Notice that the x²-4 is a difference of two squares, so can be further factorised into (x+2)(x-2)
This leaves you with a final factorisation of:
(x+2)(x-2)(x²+1)=0
Now we handle each bracket individually to obtain our four solutions for x:
x+2=0
x=-2
x-2=0
x=2
x²+1=0
x²=1
x=<span>±1</span>
Answer:
The answer is "300".
Step-by-step explanation:
Please find the complete solution in the attached file.
Answer:
n(n + 6)
Step-by-step explanation:
n² + 6n = n(n + 6)
The answer is 2, hope this helps !
Answer:
Function H h(t) = -6.1 + 19t - 55
Step-by-step explanation:
Let t = 1
<u>Function F</u>
h(t) = 6.1t^2 + 19t - 54.4
h(1) = 6.1(1)^2 + 19(1) - 54.4
h(1) = 6.1 + 19 - 54.4
h(1) = -29.3
<u>Function G</u>
h(t) = -24.1t^2 + 2.8
h(1) = -24.1(1)^2 + 2.8
h(1) = -24.1 + 2.8
h(1) = -21.3
<u>Function H</u>
h(t) = -6.1t^2 + 19t - 55
h(1) = -6.1(1)^2 + 19(1) - 55
h(1) = -6.1 + 19 - 55
h(1) = -42.1
<u>Function J</u>
h(t) = -6.1t^2 + 19t + 0.99
h(t) = -6.1(1)^2 + 19(1) + 0.99
h(t) = -6.1 + 19 + 0.99
h(t) = 13.89
