What are the solutions to this equation: 4x^2 - x + 5 = 0?
ax^2+bx+c=y quad formual is (-b +/- (sqrt b^2- 4ac))/2a, now plug in (1 +/-(sqrt 1^2- 4(4*5))/2*4 (1 +/-(sqrt 1 -80))/8 (1 +/- (sqrt -79))/8, now there isnt a sqrt of a negative meaning there are no roots, or the perlabura doesnt touch the x-axis ______________________________________... Now to find the line of symettry use -b/2a 1/8 and draw an imaginary line at (.125,0) ______________________________________... After that plug .125 into the equation for x, so 4(.125)^2- .125+5=0 or y .0625-.125+5=y -.0625+5=y 4.9375=y, so on that imaginary line plot a point at (.125,4.9375) ______________________________________... now just plug in x vaules and plot the points, since quadratics are symettrical, just count the spaces in between that point and plot another point the same number of spaces away, but on the other side of the line. please note for a perlabura to be accurate u MUST plot atleast FIVE points on the coordinate plane Hope this helps
Find the solution to the equation X2-6x+5=0?
In Yahoo!Answer, we would write it as x^2 - 6x + 5 = 0 We use the caret ^ to show an exponent (think of it as a tiny arrow pointing up) The expression on the left of the "=" is known as a "quadratic", from a Latin word (quadratum) for "square", as x^2 is often pronounced "x squared". The solutions to such an equation are often called the "zeros" of the quadratic, or even the "roots" of the quadratic. It all means the same: find the values of x that will make the quadratic add up to zero. There are many ways. One that works all the time (but it is not the easiest one) is called the "quadratic formula". You begin by writing the quadratic in this format: ax^2 + bx + c = 0 Yours is already OK, with a = 1 (we never bother to write the "1" as a coefficient) b = -6 (signs are important) c = 5 Here is the recipe: x = [ -b +/- sqrt( b^2 - 4ac ) ] / 2a where sqrt means "square root", often shown with this symbol √ (but many people find it easier to write "sqrt" in Yahoo!Answers) +/- means "plus or minus" and there will often be two answer, one with the + and one with the - b^2 - 4ac is often called the "determinant" because it determines what kind of answer you get: if b^2 - 4ac ends up being negative, you would be trying to take the square root of a negative number (which is impossible to do with real numbers). So, your problem x^2 - 6x + 5 = 0 recipe: x = [ -b +/- sqrt( b^2 - 4ac )] / 2a x = [ +6 +/- sqrt( 36 - 4(1)(5) ) ] / 2 x = [ 6 +/- sqrt( 36 - 20 ) ] / 2 x = [ 6 +/- sqrt(16) ] / 2 x = [ 6 +/- 4 ] / 2 Using the + x = (6 + 4)/2 = 10/2 = 5 Using the - x = (6 - 4)/2 = 2/2 = 1 --- As you can see, it is a long recipe method. But it is easy to learn AND it always works. If the determinant is negative, you know right away that there will be no real answer and you can stop right away. Later, when you work with other types of numbers (other than real numbers), then it will ALWAYS work, even if the determinant is negative.
Find the solution to the equation 15^2x = 3*5^(x+1) in the form log a/log b where a and b are positive integers.?
15^(2x) = 3 * 5^(x + 1) 2x * ln(15) = ln(3) + (x + 1) * ln(5) 2x * ln(15) = ln(3) + x * ln(5) + ln(5) 2x * ln(15) - x * ln(5) = ln(3) + ln(5) x * (2 * ln(15) - ln(5)) = ln(3 * 5) x * ln(15^2 / 5) = ln(15) x * ln(225 / 5) = ln(15) x * ln(45) = ln(15) x = ln(15) / ln(45)
The number of solutions of the equation 5^x+5^-x=log25/log10 (i.e log 25 to the base 10) is?
The equation hasn't any solution indeed. First, you should observe that 5^(-x)=1/(5^x), according to negative exponent properties. That way, you get 5^x + 1/(5^x) = lg 25. (where lg 25 = log25/log10, written shortly). Note t = 5^x. Resolve the equation in t variable: t + 1/t = lg 25. Multiply with t (obviously t cannot be zero). Equation comes to t^2 - lg25*t +1 = 0. That is a second degree equation. The discriminant of equation is negative: (lg 25)^2 - 4 < 0. (lg 25 < 1.4, approx.) 1.4^2 = 1.96 < 2 or 1,96^2 < 2^2 = 4. So, (lg 25)^2 < 4. Therefore, lg 25 - 4 < 0. With a negative discriminant, the equation in t variable has just complex solutions, but not real. Return to notation for t. 5^x = t. So, there is no real number x that satisfy this equation. (q.e.d)
[x] +{x} = xNow, coming to our question4{x}=[x]+xSubstituting the value of x,4{x}=[x]+[x]+{x}Subtracting {x} from both sides,3{x}=2[x]Dividing both sides by 3,{x}=2[x] /3Now, as we know that0<= {x} < 10<= 2[x]/3 < 10<= [x] < 3/2So, solution of above is[x] = 0 or [x] = 1Case (i) if [x] = 1,{x} = 2[x] /3= 2/3As, x =[x] + {x}x=1+2/3 =5/3Case (ii) if [x] =0{x} =2[x]/3= 0As, x = [x] +{x}x=0+0 =0So, solutions of4{x}=[x]+x are x=0 and x = 5/3.
We could write from top to bottom like this:5^(x-1) = 5^x - 55^x * 5^(-1) = 5^x - 5(5^x)/5 = 5^x - 55^x = 5*(5^x - 5)5^x = 5*5^x - 2525 = 5*5^x - 5^x (As moving 25 to the left side and 5^x to the right side)25 = 5^x * (5–4)25 = 4 * 5^x5^x = 25/4 = 6.25log(5^x) = log(6.25) (Taking the natural log of both sides)x*log(5) = log(6.25) (The exponent inside the log as the x can be taken out front as a multiplier)x = log(6.25)/log(5)x = 1.138646884
First, multiply both sides by the common denominator to eliminate the fractions:(2x+6)/(x-5) = (x+3)(x-1)The common denominator is (x-5)(x-1):(2x+6)(x-1) = (x+3)(x-5)Use the FOIL method on each set of parentheses:2x^2–2x+6x-6 = x^2–5x+3x-8Combine like terms:2x^2–2x+6x-6 = x^2–5x+3x-82x^2+4x-6 = x^2–2x-8As the highest exponent in this equation is equal to 2 it is a quadratic equation and must be formatted like one. To do this we’ll need to subtract all of the terms from one side of the equation from both sides.2x^2+4x-6 = x^2–2x-8Subtract (x^2–2x-8) from both sides:x^2+6x+2 = 0Normally from here you would factor this equation. You would need two factors of 2 such that, when multiplied together, the product is 2 but added together yield a sum of 6. There are no factors that fit this definition, which means we need to use the quadratic formula to solve this.x^2+6x+2 = 0x = (-b +/-(square root)b^2–4ac)/2aa = 1b = 6c = 2x = (-6+/-(square root)6^2–4*2)/2x = (-6+/-(square root)36–8)/2x = (-6+/-(square root)28)/2We’ll approximate the square root of 28 as 5.292:x = (-6+/-5.292)/2x = (-11.29)/2x = (-5.645)ORx = (-1.134)/2x = -0.567Now substitute to prove this:x^2 + 6x + 2 = 0x = -5.645(-5.645)^2 + 6(-5.645) + 2 = 031.3+ (-33.87) + 2 = 0(-2.57) + 2 = 00.57 APPROXIMATELY = 0So this solution works! Now how about the other one?x^2 + 6x + 2 = 0x = -0.567(-0.567)^2 + 6(-0.567) + 2 = 00.312489 - 3.402 + 2 = 0-3.089511 + 2 = 0-1.085911 approximately = 0So there you have it! Both of your solutions for x. Remember that we used approximate values for the square root of 28 earlier which resulted in our values being slightly off. This is to be expected when using approximate values for non-terminating decimals.
3X/5 – 2 = X + 1/3 Solve the equation above for x.?
how many times are you going to ask this? learn to use parans, this is very ambiguous first guess (3X/5) – 2 = X + (1/3) multiply by 15 9X – 30 = 15X + 5 9X – 15X = 5 + 30 –6X = 35 X = –35/6
Consider any polynomial equation with integer coefficients:[math]a_n x^n + \cdots + a_1 x + a_0 = 0[/math].Let's assume that x is rational. Then we can write x in the form p/q, where p and q are relatively prime integers with q non-zero. Let's plug that into the equation:[math]a_n (p/q)^n + \cdots + a_1 (p/q) + a_0 = 0[/math].To get rid of fractions, let's multiply both sides by the non-zero value [math]q^n[/math]. This is the result:[math]a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0[/math].You may notice that all terms except for [math]a_0 q^n[/math] are multiples of p. Hence, [math]a_0 q^n[/math] must be a multiple of p, too. But as p and q are relatively prime, this means that p has to divide [math]a_0[/math]. For the same reason, [math]q[/math] has to divide [math]a_n[/math].In your specific case, if there is any rational solution p/q then we know that q must divide 1 and p must divide -10. This gives us only eight candidates for a rational solution:[math]\{-10,-5,-2,-1,1,2,5,10\}[/math]Check all eight manually, see that none of them are roots of your equation, and you are done.