Find all zeros of f(x) = 2x^4 - 5x^3 +3x^2 + 4x - 6?
2x^4 - 5x^3 +3x^2 + 4x - 6 2,5,3,4,6=20 even means +/-1 might work x=1 2-5+3+4-6=not zero x= -1 [only the odd exponents flip signs] 2+5+3-4-6=0 x= -1 means the factor is (x+1) synthetic division for 2,-5,+3,+4,-6 take the coefficient of the first term 2 multiply by the value of x -1(2)= -2 ADD to the next term -7 repeat -1(-7)=7 7+3=10 10 -1(10)= -10 -10+4= -6 -6 -1(-6)=6 -6+6=0 collect the coefficients 2,-7,+10,-6 gives you 2x^3-7x^2+10x-6 does not give 'zero' for multiples of 6 [+/-1,+/-2,+/-3,+/-6] so fractions 1/2 is the only denominator +/-3/2,+/-1/2 [6/2 is +/-3; 2/2 is +/-1] x= 3/2 2x^3-7x^2+10x-6 2(27/8)-7(9/4)+10(3/2)-6 (27/4)-(63/4)+(60/4)-(24/4)=0 x=3/2 extract the factor 2x=3 2x-3=0 (2x-3) long division (2x^3-7x^2+10x-6)/(2x-3) 2x^3/2x=x^2 x^2(2x-3)=2x^3-3x^2 subtract -4x^2+10x-6 -4x^2/2x= -2x -2x(2x-3)= -4x^2+6x subtract 4x-6 4x/2x=2 2(2x-3)=4x-3 collect the multiples x^2-2x+2 complete the square x^2-2= -2 x^2-2x+1= -2+1 (x-1)^2= -1 x-1= +/-1i x= 1+/-i altogether x= -1,3/2,(1+/-i) you give points?
The equation x² + 10x + k = 0 has equal roots, find the value of k?!?
k=25 the roots are 5,5 In general for any eqn ax^2 +bx+c=0 the roots are -b + sqrt(b^2-4ac)/2a , -b - sqrt(b^2-4ac)/2a The simple way to do this is by equating the b^2-4*a*c=0 since they are equal. so then 10^2-4*1*k=0 Solve to get k=25 You can double check by putting 25 and finding the roots which are 5,5 and hence they are equal!!
What are the roots of the equation [math] x^4 - 4x^3 + 8x^2 - 8x + 4 =0 [/math] ?
x^4-4x^3+8x^2-8x+4 =0 can be written as x^4-4x^3+(4x^2+4x^2)-8x+4=0 {8x^2=4x^2+4x^2}Now group the part of the equation in this manner.x^2(x^2-4x+4)+4(x^2-2x+1)=0x^2(x-2)^2+4(x-1)^2=0we can see that (x(x-2))^2 and (2(x-1))^2 would always be positive.So to get the sum equal to zero we need both the parts to be equal to zero simultaneously. But this cannot be possible in this case so real roots do not exist in this case, we have to find imaginary roots.x^2(x-2)^2=-4(x-1)^2(x^2(x-2)^2)/(x-1)^2=-4((x^2-2x)/x-1)^2=-4((x^2-2x)/x-1)=+-2isolve for this equations
Find the set of values of k for which the equation 2x^2+kx+2+0 has no real roots ?
Find the set of values of k for which the equation 2x^2+kx+2+0 has no real roots ? When using the quadratic formula to solve a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. ) For a = 2, b = k and c =2, b^2 - 4ac = k^2 – 16 If this is < 0, then k^2 < 16 and k < 4 But k > -4 also satisfies the inequality, hence -4 < k < 4
If x^2 + kx + k has two distinct real solution, then the value of k will satisfy?
You need write x^2+ka+k = 0 has 2 distinct real roots. Only an equation has roots & an expression like x^2+kx+k does not have any roots. Now for amswer -The discriminant b^2–4ac of the equation is k^2–8k. This must be > 0So k(k-8) >0. It can happen if both the terms k, k-8 are < 0 OR when both are >0Both are less than zerok<0 and k-8<0 => k<0 and k<8 or k<0 (lower of the values)Both are > 0k>0 and k-8>0 => k>8The result is when k <0 or k>8 the roots are real and distinct
How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?
The equation of the tangent to the curve [math]y=f(x)[/math] that passes the point [math](x_0 , y_0 )[/math] can be determined by writing down general expression of tangent equation at [math] (k, f(k)) [/math], then determining [math]k [/math] using the fact that the tangent line should pass the point [math](x_0 , y_0 )[/math].The reason we use the tangent equation at [math] (k, f(k)) [/math] to solve the problem is because it not only covers all the tangent equation of the curve, but it is fairly easy to compute since we know the tangent value of the line( which is [math]f'(k)[/math]) and one point that the line passes( [math] (k, f(k)) [/math] ). Because of this, this method is also useful when solving any problems that asks you to determine a tangent equation of a curve that qualifies a certain property.So, assuming [math]k[/math] as the [math]x[/math] coordinate of the point that the tangent line meets with the curve, we obtain the equation of the tangent line.[math] y = f'(x)( x-k) +f(k) [/math][math]= (2-2k)(x-k) + 2k - k^2[/math][math] = (2-2k)x +k^2 [/math]Since this equation should pass the point (2,9), we determine the value of k.[math] 9 = (2-2k)*2 + k^2 [/math][math] k^2 -4k -5 = (k-5)(k+1) = 0 [/math][math] k = 5 or -1 [/math]Thus, [math] y = 4x+1 or -8x+25 [/math]I have not studied mathematics in English, so if I have used any awkward expression of any kind, please do not hesitate to edit my answer.
How can the quadratic equation [math]x^2 - 3x - 10 = 0[/math] be solved?
You can use the quadratic formula, as others pointed out, or you can factorize the formula. ( I recommend using the quadratic formula, because it is simple and always works.)Factorization involves the following steps.Find two numbers whose sum is -3 and product is -10. (In general, for [math]ax^2 + bx + c[/math], find two numbers whose sum is b and the product is ac.) With some thought, you identify that they are -5 and 2.Write [math]x^2 - 3x - 10[/math] as [math]x^2 - 5x + 2x -10[/math].Now, find the common factor in the first two terms and last two terms. [math]x(x-5) + 2 (x-5)[/math].Now, find the common factor in these two terms. [math](x-5)(x+2)[/math]So, the equation becomes [math](x-5)(x+2) = 0[/math]which means either [math](x-5) = 0[/math] or [math](x+2) = 0[/math].So, x = 5 or x = -2.Note that you can put the two terms in either order. [math]\begin{align}x^2 - 3x - 10 &= x^2 + 2x - 5x - 10 \\ &= x(x+2) - 5(x+2) \\ &= (x+2)(x-5)\end{align}[/math].We can solve a little more complex quadratic equations also this way. For example, solve[math]10x^2 - x - 3 = 0[/math] Here, we need to find two numbers whose sum is -1 and product is [math]10 \cdot -3 = -30[/math]. We find they are -6 and 5. So,[math]\begin{align}10x^2 - x - 3 &= 10x^2 - 6x + 5x - 3 \\ &= 2x(5x-3) + 1(5x - 3) \\ &= (2x+1)(5x-3)\end{align}[/math]So, x is -1/2 or 3/5.However, this method has a problem: We may not find those two numbers whose sum and product are the given numbers easily. (And trying to find those mathematically will gives us the same quadratic equation!) We can solve the equation using quadratic formula as easy as this, and it works. Let us see.The solution to [math]ax^2 + bx + c = 0[/math] is[math]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/math]First problem:[math]x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot -10}}{2 \cdot 1} = \frac{3 \pm 7}{2} = 5[/math] or [math]-2[/math].[math]x = \frac{1 \pm \sqrt{1- 4 \cdot 10 \cdot 3}}{2 \cdot 10} = \frac{1\pm 11}{20} = \frac{3}{5}[/math] or [math]-\frac{1}{2}[/math].If we have a quadratic equation like [math]x^2-x-10 = 0[/math], we cannot factorize the expression easily, so will have to use the quadratic formula.Moral: Always use the quadratic formula.
For what "k" the quadratic equation kx^2-5x+k=0 has equal roots?
Bold Question: a quadratic equation has equal roots when discriminant is zero. B^2 - 4AC = 0 In this case, A = k, B = -5, and C = k We need to solve (-5)^2 - 4(k)(k) = 0 25 - 4k^2 = 0 -4k^2 = -25 k^2 = 25/4 k = +/- 5/2 Question 1: Let's find the roots using quadratic formula x = (-(-5) +/- sqrt((-5)^2 - 4(3)(2))) / (2*3) x = (5 +/- sqrt(25 - 24)) / 6 x = (5 +/- sqrt(1)) / 6 x = (5 +/- 1) / 6 x = (5+1) / 6 or x = (5-1) / 6 x = 1 or x = 2/3 Solution: x = 1 or x = 2/3 Question 2: There is no k in this equation so you cannot find it Question 3: Show that the discriminant is zero. Find B^2 - 4AC where A = 2, B = -4, C = 3. (-4)^2 - 4(2)(3) 16 - 24 -8 This isn't zero. So it does not have equal roots. Question 4: Find the roots first and then their sum and product. x = (-(5) +/- sqrt((-5)^2 - 4(2)(-4))) / (2*2) x = (5 +/- sqrt(25 - -32)) / 4 x = (5 +/- sqrt(57)) / 4 x = (5 + sqrt(57)) / 4 or x = (5 - sqrt(57)) / 4 Sum: [(5 + sqrt(57)) / 4] + [(5 - sqrt(57) / 4] = [5 + sqrt(57) + 5 - sqrt(57)] / 4 = 10/4 = 5/2 Product: [(5 + sqrt(57)) / 4] * [(5 - sqrt(57) / 4] = [25 - 5sqrt(57) + 5sqrt(57) - 57] / 16 = -32/16 = -2 Question 5: Need to find x and x + 1 such that x^2 + (x + 1)^2 = 25 x^2 + x^2 + 2x + 1 = 25 2x^2 + 2x + 1 = 25 2x^2 + 2x - 24 = 0 x^2 + x - 12 = 0 (x + 4)(x - 3) = 0 x = -4 or x = 3 We want positive numbers so we can ignore x = -4 x = 3. Then x + 1 = 4 Solution: 3,4
What are the different ways to solve quadratic equations?
To understand the different methods, let us take an example.Solving: [math]x^2 - 7x + 12 = 0[/math]Factoring[math]x^2 - 7x + 12 = 0[/math][Splitting the middle term] [math]x^2 - 3x - 4x + 12 = 0[/math][Taking terms pairwise][math]x(x-3) -4(x-3) = 0[/math][Pulling out the common term][math](x-3)(x-4) = 0[/math]Or,[math]x = 3 [/math]or [math]x = 4[/math] are roots of the equation.Completing the squareThe general principle requires the generic quadratic equation [math]ax^2 + bx + c = 0[/math] be re-written in the form a complete square.[math]x^2 - 7x + 12 = 0[/math] can be re-written as[math]x^2 - 2(x)(7/2) + (7/2)^2 + 12 = (7/2)^2 [/math][math](x - 7/2)^2 = 49/4 - 12 [/math][math]\implies (x - 7/2)^2 = 1/4[/math]The roots are [math]x = +1/2 + 7/2 = 4 [/math] or [math]-1/2 + 7/2 = 3[/math]Quadratic formulaThe general formula is given by [math]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]For [math]x^2 - 7x + 12 = 0, x = \frac{+7\pm\sqrt{7^2-4(1)(12)}}{2} [/math][math]\implies x = \frac{7 \pm \sqrt{1}}{2} [/math][math]\implies x = \frac{7 \pm 1}{2}[/math]The roots are [math]x = 4[/math] or [math]x = 3[/math]Graphing the EquationsGraphing a quadratic equation results in a parabola, with the zeroes indicated by the intersection of the curve with the x-axis.The graph of [math]x^2 - 7x + 12[/math] intersects x-axis at 3 & 4 respectively.
Calculus AP Free Response: Let f be the function defined by f(x)=(x^2-3) e^x for all real numbers x.?
A) f(x)=(x^2-3) e^x f'(x) = e^x (x^2-3) + e^x (2x) = 0 e^x ( x^2-2x-3)=0 e^x ( x-3)(x+1)=0 x=3, x=-1 Plot the points on a number line: ----------------------- -∞'''''''''''-1''''''3''''''∞ Consider the intervals (-∞,-1), (-1,3), and (3, ∞) Choose 'any' value from each interval Evaluate f'(x) = e^x ( x^2-2x-3) in each of these intervals If f'(x) < 0, f(x) is decreasing on that interval. If f'(x) > 0, f(x) is increasing on that interval. (-∞, -3): choose -4 ; e^x ( x^2-2x-3) = 57.08 > 0, so f is increasing on (-∞, -3) (-3,1): Choose 0: e^x ( x^2-2x-3) = -8.15 < 0, so f is decreasing on (-3, 1) (3,∞) : choose 4 : e^x ( x^2-2x-3) = 13.6 > 0, so f is increasing on (3,∞) f is increasing on (-∞, -3) U (3,∞) : B) f'(x) = e^x (x^2-3) + 2x e^x f''(x) = e^x (x^2-3) + 2x e^x + 2 e^x + 2xe^x = 0 e^x ( x^2-3+2x+2+2x) = 0 e^x (x^2+4x-1) = 0 The points of inflection are the roots of the equation x^2+4x-1=0 This equation is of form ax^2+bx+c=0 a = 1 b = 4 c = -1 x=[-b+/-sqrt(b^2-4ac)]/2a] x=[-4 +/-sqrt(4^2-4(1)(-1)]/(2)(1) discriminant is b^2-4ac =20 x=[-4 +√(20)] / (2)(1) x=[-4 -√(20)] / (2)(1) x=[-4+4.47213595499958] / 2 x=[-4-4.47213595499958] / 2 The roots are 0.2361 and -4.2361 f(0.2361) = -8.0033 f(-4.2361) = 40.62 (0.2361,-8.003) and (-4.2361,40.62) C) f''(x) = e^x (x^2+4x-1) when x=-3, f''(x) < 0, so f has a maximum when x=1, f''(x) > 0, so f has a minimum.