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What are the roots of x squared plus 8x plus 25?
We will use a tool called a quadratic formula. First, I will prove it. For all degree two function f(x) = ax^2 + bx + c where a, b, c are constants, claim the roots are -b +/- sqrt(b^2 - 4ac))/2a IF a is not 0 and b^2 - 4ac >= 0. Proof. Suppose the conditions are satisfied, finding the roots of the function is the same as solving the following ax^2 + bx + c = 0 we will simplify, since a != 0 a(x^2 + (b/a)x) + c = 0 we will do a trick, a trick that you see once, you will never forget, we add then subtract the same thing a(x^2 + (b/a)x + b^2/(4a^2) - b^2/(4a^2)) + c = 0 Now notice b/a = 2 . b/2a, we will do that, also, we will take -b^2/(4a^2) out, notice, b^2/4a^2 = (b/2a)^2 Then we get a (-b^2/(4a^2)) + a( x^2 + 2 (b/2a)x + (b/2a)^2) + c = 0 we use the complete square identity -b^2/(4a) + a(x + b/2a)^2 + c = 0 Isolate x a(x + b/2a)^2 = -c + b^2/4a = (-4ac + b^2)/4a (x + b/2a)^2 = (b^2 - 4ac)/(4a^2) (x + b/2a) = +/- sqrt (b^2 - 4ac) / 2a x = +/- sqrt (b^2 - 4ac) / 2a - b/2a x = -b +/- sqrt(b^2 - 4ac))/2a Q.E.D Now apply the quadratic formula to your question: x^2 + 8x + 25. First check if a = 0, a = 1 != 0, pass. Now check if b^2 - 4ac is non-negative. 8^2 - 4 x 1 x 25 = 64 - 100 = - 36 < 0. Done, this function have no roots.
What is a quadratic equation?
A quadratic equation is an equation where a quadratic polynomial is equal to zero. It can be written as ax^2+bx+c=0 where a,b,c are the coefficients and x is the variable. A quadratic equation has always two complex solutions for x given by the formula x=-b/2a+sqrt(b^2-4ac)/2a and x=-b/2a-sqrt(b^2-4ac)/2a. Examples of quadratic equations are x^2+x-2=0, 5x^2+6x=0, x^2+1=0 etc.
Ax2 plus bx plus c equals 0?
x=(-b+sqrt(b^2-4*a*c))/(2a) and (-b-sqrt(b^2-4*a*c))/(2a) (This is called the quadratic formula)
How do you solve for the variable b in 2a plus 2b equals c?
2a + 2b = c subtract 2a from both sides 2a - 2a + 2b = c - 2a 2b = c - 2a divide both sides by 2 (2/2)b = (c - 2a)/2 b = (c - 2a)/2 --------------------
How do you find the max and min in a quadratic equation?
A quadratic function is of the form: f(x) = ax2 + bx + c where a ≠0 If a > 0 then the quadratic has a minimum but its maximum, asymptotically, is +∞. If a < 0 then the quadratic has a maximum but its minimum, asymptotically, is -∞. The extremum (whichever it is) is attained when x = -b/2a. The extreme value is f(-b/2a) = a*(-b/2a)2 + b(-b/2a) + c = b2/4a - b2/2a + c = -b2/4a + c
How many roots does the polynomial have 4x2 - 3x - 5?
4x**2-3x-5 quadratic = (-b+sqrt(b**2-4ac))/2a quadratic = (-b-sqrt(b**2-4ac))/2a b**2-4ac>0 --> 2 roots b**2-4ac=0 --> 1 root b**2-4ac<0 --> 0 roots (-3)**2-4(4*-5) 9+80=89 2 roots = (6+89**0.5)/8, (6-89**0.5)/8
Who can factor 64a6n - b6n?
n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)
Where did the quadratic formula come from?
The quadratic formula originated from the concept of completing the square. let's take ax2 + bx + c = 0. To complete the square, solve for x. Subtract c. ax2 + bx = -c. Then divide by a [notice- if there is no a value, then a=1]. x2 + bx/a = -c/a. Add (b/2a)2 to both sides. x2 + bx/a + b2/4a2 = -c/a + b2/4a2 Factor/Reformat. (x + b/2a)2 = (b2-4ac) / 4a2 (x + b/2a)2 = [(b2-4ac) / 2a]2 Square-root both sides. x + b/2a = ± √(b2-4ac) / 2a Subtract b/2a. x = -b/2a ± √(b2-4ac) / 2a Combine terms. x = [-b ± √(b2-4ac)] / 2a
Where did the quadratic equation come from?
The quadratic formula originated from the concept of completing the square. let's take ax2 + bx + c = 0. To complete the square, solve for x. Subtract c. ax2 + bx = -c. Then divide by a [notice- if there is no a value, then a=1]. x2 + bx/a = -c/a. Add (b/2a)2 to both sides. x2 + bx/a + b2/4a2 = -c/a + b2/4a2 Factor/Reformat. (x + b/2a)2 = (b2-4ac) / 4a2 (x + b/2a)2 = [(b2-4ac) / 2a]2 Square-root both sides. x + b/2a = ± √(b2-4ac) / 2a Subtract b/2a. x = -b/2a ± √(b2-4ac) / 2a Combine terms. x = [-b ± √(b2-4ac)] / 2a
Which of the following conditions would make 2a - 2b 0?
If a=b or both a and b =0
GIven Tx equals ax2 plus bx plus c find a b and c if T0 equals -4 T1 equals -2 and T2 equals 6?
T(x) = ax^2 + bx + cT(0) = -4T(1) = -2T(2) = 6Solution:Since T(0) = -4, then c = -4.So, when x = 1, we have:-2 = a(1)^2 + b(1) - 4-2 = a + b - 4-2 + 4 = a + b - 4 + 42 = a + b2 - a = a - a + b2 - a = bWhen x = 2, we have:6 = a(2)^2 + b(2) - 46 = 4a + 2b - 46 + 4 = 4a + 2b - 4 + 410 = 4a + 2b10/2 = 4a/2 + 2b/25 = 2a + b5 - 2a = 2a - 2a + b5 - 2a = b2 - a = 5 - 2a2 - 2 - a + 2a = 5 - 2 - 2a + 2aa = 32 - a = b2 - 3 = b-1 = bThus, a = 3, b = -1, and c = -4.Check:
Why does the quadratic formula work every time for quadratic equations?
The quadratic formula works for every quadratic equation because it is the standard form of a quadratic solved for x. ax2+bx+c=0 - Standard Form, how quadratic equations are normally displayed x2+bx/a+c/a=0 - Divide both sides by a, Division Property of = x2+(b/a)x= -(c/a) - Separate variables and constants, Subtraction Property of = x2+(b/a)x+(b/2a)2=(b/2a)2-(c/a) - Complete the square. (b/2)2 (x+b/2a)2=(b2-4ac)/4a2 - Factor and Simplify x+b/2a=+/-sqrt((b2-4ac)/4a2) - Square Root both sides. x=-(b/2a)+/-sqrt(b2-4ac)/2a - Solve for x, Subtraction Property of = x=(-b+/-sqrt(b2-4ac))/2a - The Quadratic Equation, Simplify
