💡Okay, so I have this problem where two cars leave town at the same time, going in opposite directions. One car is going at 70 miles per hour, and the other is going at some speed that I need to find out. After 2 hours, they're 250 miles apart. Hmm, okay, let's see how to approach this.First, I remember that when two objects move in opposite directions, their speeds add up when calculating the distance between them. So, if one car is going at 70 mph and the other at, say, 'v' mph, then together, they're moving away from each other at a combined speed of 70 + v mph.Now, distance is equal to speed multiplied by time. In this case, the time is 2 hours, and the total distance between them is 250 miles. So, I can set up an equation where the sum of the distances each car travels equals 250 miles.Let me write that down:Distance by first car + Distance by second car = Total distance apart70 mph * 2 hours + v mph * 2 hours = 250 milesCalculating the first part, 70 mph times 2 hours is 140 miles. So, the equation becomes:140 miles + 2v = 250 milesNow, I need to solve for 'v'. Subtracting 140 miles from both sides gives:2v = 250 - 1402v = 110Then, dividing both sides by 2:v = 110 / 2v = 55So, the speed of the second car is 55 mph. Let me double-check that. If one car goes 70 mph for 2 hours, it covers 140 miles. The other car at 55 mph for 2 hours covers 110 miles. Adding those together, 140 + 110 equals 250 miles, which matches the given total distance. So, that makes sense.I think I've got it right. The second car must be traveling at 55 miles per hour.

💡Alright, so I have this problem about mappings and whether they have a certain property called property p. Let me try to understand what property p is first. The problem says that a mapping f from V to the real numbers has property p if for any two vectors a and b in V, and any real number λ, the following holds:f[λa + (1 - λ)b] = λf(a) + (1 - λ)f(b)Okay, so this looks familiar. It seems like it's talking about linearity. If I remember correctly, a function is linear if it satisfies two properties: additivity and homogeneity. Additivity means that f(a + b) = f(a) + f(b), and homogeneity means that f(λa) = λf(a). But wait, the definition here is a bit different. It's not exactly additivity and homogeneity, but it's a combination of both. Let me think. If I set λ = 1, then f[a] = f(a), which is trivial. If I set λ = 0, then f[b] = f(b), which is also trivial. So, it's more like a convex combination. So, property p is about the function preserving convex combinations. That makes sense because in convex functions, the function evaluated at a convex combination is less than or equal to the convex combination of the function values. But here, it's an equality, so it's not just convex, it's affine.Wait, affine functions are functions that are linear plus a constant. So, maybe property p is equivalent to the function being affine. Let me check. If f is affine, then f(λa + (1 - λ)b) = λf(a) + (1 - λ)f(b). Yes, that's exactly the definition of an affine function. So, property p is equivalent to the function being affine. So, now I need to check whether each of the given mappings is affine. Let's look at each one.1. f₁(m) = x - y, where m = (x, y). Hmm, this is a linear function because it's just a linear combination of x and y. Since linear functions are also affine (with the constant term being zero), f₁ should have property p.2. f₂(m) = x² + y, where m = (x, y).This one is a bit trickier. It has x squared, which is a nonlinear term. So, I suspect this is not affine. Let me test it with specific vectors. Let me choose a = (0, 0) and b = (1, 2). Then, let's compute f₂(λa + (1 - λ)b) and see if it equals λf₂(a) + (1 - λ)f₂(b).First, λa + (1 - λ)b = (0, 0) + (1 - λ)(1, 2) = (1 - λ, 2(1 - λ)). So, f₂(λa + (1 - λ)b) = (1 - λ)² + 2(1 - λ) = 1 - 2λ + λ² + 2 - 2λ = λ² - 4λ + 3.On the other hand, λf₂(a) + (1 - λ)f₂(b) = λ(0² + 0) + (1 - λ)(1² + 2) = 0 + (1 - λ)(3) = 3 - 3λ.So, f₂(λa + (1 - λ)b) = λ² - 4λ + 3, and λf₂(a) + (1 - λ)f₂(b) = 3 - 3λ. These are not equal unless λ² - 4λ + 3 = 3 - 3λ, which simplifies to λ² - λ = 0. This holds only when λ = 0 or λ = 1, but not for all real numbers λ. Therefore, f₂ does not have property p.3. f₃(m) = x + y + 1, where m = (x, y).This looks like an affine function because it's a linear combination of x and y plus a constant. Let me verify it formally.Compute f₃(λa + (1 - λ)b). Let a = (x₁, y₁) and b = (x₂, y₂). Then,λa + (1 - λ)b = (λx₁ + (1 - λ)x₂, λy₁ + (1 - λ)y₂).So, f₃(λa + (1 - λ)b) = [λx₁ + (1 - λ)x₂] + [λy₁ + (1 - λ)y₂] + 1 = λ(x₁ + y₁) + (1 - λ)(x₂ + y₂) + 1.Now, compute λf₃(a) + (1 - λ)f₃(b):λf₃(a) = λ(x₁ + y₁ + 1) = λ(x₁ + y₁) + λ.(1 - λ)f₃(b) = (1 - λ)(x₂ + y₂ + 1) = (1 - λ)(x₂ + y₂) + (1 - λ).Adding them together: λ(x₁ + y₁) + (1 - λ)(x₂ + y₂) + λ + (1 - λ) = λ(x₁ + y₁) + (1 - λ)(x₂ + y₂) + 1.Which is the same as f₃(λa + (1 - λ)b). Therefore, f₃ satisfies property p.So, summarizing:- f₁ is affine, so it has property p.- f₂ is not affine because of the x² term, so it doesn't have property p.- f₃ is affine, so it has property p.I think that's it. Let me just double-check if I made any mistakes. For f₂, I chose specific vectors and showed that the equality doesn't hold for all λ, which should be sufficient. For f₁ and f₃, I verified the property holds in general, so they should be fine.**Final Answer**The mappings that have property ( p ) are boxed{1} and boxed{3}.

💡Alright, so I have this geometry problem here. Let me try to understand it step by step. We have a triangle ABE, and points C and D are on side BE such that BC = CD = DE. So, BE is divided into three equal parts. That means C is one-third of the way from B to E, and D is two-thirds of the way. Got it.Now, we have four points: X, Y, Z, and T. These are the circumcenters of triangles ABE, ABC, ADE, and ACD, respectively. I need to prove that T is the centroid of triangle XYZ. Hmm, okay. First, let me recall what a circumcenter is. The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. It's also the center of the circumscribed circle around the triangle. So, for each of these triangles, X, Y, Z, and T are the centers of their respective circumscribed circles.Now, the centroid of a triangle is the point where the three medians intersect. A median is a line segment from a vertex to the midpoint of the opposite side. The centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex.So, to show that T is the centroid of triangle XYZ, I need to show that T lies at the intersection of the medians of triangle XYZ and that it divides each median in a 2:1 ratio.Let me try to visualize this. Maybe drawing a diagram would help. Since I can't draw here, I'll try to imagine it. Triangle ABE with BE divided into three equal parts by C and D. Then, triangles ABC, ADE, and ACD are formed. Each of these has a circumcenter, which are points Y, Z, and T respectively. X is the circumcenter of the main triangle ABE.I think it might help to consider the properties of circumcenters and centroids. Since circumcenters are related to perpendicular bisectors, maybe I can find some relationships between the perpendicular bisectors of these smaller triangles and the medians of triangle XYZ.Let me think about the positions of these circumcenters. For triangle ABE, the circumcenter X is somewhere outside or inside depending on the type of triangle. Similarly, Y, Z, and T will be located based on the specific triangles ABC, ADE, and ACD.Since BC = CD = DE, the points C and D are equally spaced along BE. Maybe this symmetry can help in showing that T is the centroid.Perhaps I can use coordinate geometry. Assign coordinates to the points and calculate the positions of X, Y, Z, and T. Then, check if T is indeed the centroid by verifying the centroid formula.Let me assign coordinates to the points. Let's place point B at (0, 0), E at (3, 0), so that BE is along the x-axis with length 3 units. Then, since BC = CD = DE, each segment is 1 unit. So, point C is at (1, 0), and point D is at (2, 0).Now, let me assign coordinates to point A. Since the problem doesn't specify where A is, I can choose it arbitrarily. Let's say point A is at (a, b). This way, I can keep it general.So, points are:- B: (0, 0)- C: (1, 0)- D: (2, 0)- E: (3, 0)- A: (a, b)Now, I need to find the circumcenters X, Y, Z, and T.Starting with X, the circumcenter of triangle ABE. The circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle.Let me find the perpendicular bisector of AB and the perpendicular bisector of AE.First, find the midpoint of AB. Midpoint M1 is ((a + 0)/2, (b + 0)/2) = (a/2, b/2).The slope of AB is (b - 0)/(a - 0) = b/a. Therefore, the slope of the perpendicular bisector is -a/b.So, the equation of the perpendicular bisector of AB is:(y - b/2) = (-a/b)(x - a/2)Similarly, find the midpoint of AE. Midpoint M2 is ((a + 3)/2, (b + 0)/2) = ((a + 3)/2, b/2).The slope of AE is (b - 0)/(a - 3) = b/(a - 3). Therefore, the slope of the perpendicular bisector is -(a - 3)/b.So, the equation of the perpendicular bisector of AE is:(y - b/2) = [-(a - 3)/b](x - (a + 3)/2)Now, the circumcenter X is the intersection of these two perpendicular bisectors. Let me solve these two equations to find X.First equation:y = (-a/b)(x - a/2) + b/2Second equation:y = [-(a - 3)/b](x - (a + 3)/2) + b/2Set them equal:(-a/b)(x - a/2) + b/2 = [-(a - 3)/b](x - (a + 3)/2) + b/2Subtract b/2 from both sides:(-a/b)(x - a/2) = [-(a - 3)/b](x - (a + 3)/2)Multiply both sides by b to eliminate denominators:-a(x - a/2) = -(a - 3)(x - (a + 3)/2)Simplify signs:a(x - a/2) = (a - 3)(x - (a + 3)/2)Expand both sides:Left side: a*x - (a^2)/2Right side: (a - 3)*x - (a - 3)*(a + 3)/2Simplify right side:(a - 3)*(a + 3) = a^2 - 9, so right side becomes (a - 3)x - (a^2 - 9)/2So, equation is:a*x - (a^2)/2 = (a - 3)x - (a^2 - 9)/2Bring all terms to left side:a*x - (a^2)/2 - (a - 3)x + (a^2 - 9)/2 = 0Factor x terms:[a - (a - 3)]x + [ - (a^2)/2 + (a^2 - 9)/2 ] = 0Simplify:[3]x + [ (-a^2 + a^2 - 9)/2 ] = 0Which simplifies to:3x - 9/2 = 0So, 3x = 9/2 => x = 3/2Now, substitute x = 3/2 into first equation to find y:y = (-a/b)(3/2 - a/2) + b/2Simplify:y = (-a/b)( (3 - a)/2 ) + b/2 = [ -a(3 - a) ] / (2b) + b/2So, y = [ -3a + a^2 ] / (2b) + b/2Combine terms:y = (a^2 - 3a)/ (2b) + b/2To combine, let's get a common denominator:y = (a^2 - 3a + b^2) / (2b)So, circumcenter X is at (3/2, (a^2 - 3a + b^2)/(2b))Okay, that's X. Now, let's find Y, the circumcenter of triangle ABC.Points of triangle ABC: A(a, b), B(0,0), C(1,0)Again, find perpendicular bisectors of AB and AC.Midpoint of AB is (a/2, b/2), same as before.Slope of AB is b/a, so perpendicular bisector slope is -a/b.Equation: y - b/2 = (-a/b)(x - a/2)Midpoint of AC: ((a + 1)/2, (b + 0)/2) = ((a + 1)/2, b/2)Slope of AC: (b - 0)/(a - 1) = b/(a - 1)Perpendicular bisector slope: -(a - 1)/bEquation: y - b/2 = [-(a - 1)/b](x - (a + 1)/2)Find intersection of these two perpendicular bisectors.First equation: same as before, y = (-a/b)(x - a/2) + b/2Second equation: y = [-(a - 1)/b](x - (a + 1)/2) + b/2Set equal:(-a/b)(x - a/2) + b/2 = [-(a - 1)/b](x - (a + 1)/2) + b/2Subtract b/2:(-a/b)(x - a/2) = [-(a - 1)/b](x - (a + 1)/2)Multiply both sides by b:-a(x - a/2) = -(a - 1)(x - (a + 1)/2)Simplify signs:a(x - a/2) = (a - 1)(x - (a + 1)/2)Expand both sides:Left: a*x - (a^2)/2Right: (a - 1)*x - (a - 1)*(a + 1)/2Simplify right side:(a - 1)*(a + 1) = a^2 - 1, so right side becomes (a - 1)x - (a^2 - 1)/2Equation:a*x - (a^2)/2 = (a - 1)x - (a^2 - 1)/2Bring all terms to left:a*x - (a^2)/2 - (a - 1)x + (a^2 - 1)/2 = 0Factor x terms:[a - (a - 1)]x + [ - (a^2)/2 + (a^2 - 1)/2 ] = 0Simplify:[1]x + [ (-a^2 + a^2 - 1)/2 ] = 0Which is:x - 1/2 = 0 => x = 1/2Substitute x = 1/2 into first equation:y = (-a/b)(1/2 - a/2) + b/2 = (-a/b)( (1 - a)/2 ) + b/2 = [ -a(1 - a) ] / (2b) + b/2Simplify:y = (a^2 - a)/(2b) + b/2Combine terms:y = (a^2 - a + b^2)/(2b)So, circumcenter Y is at (1/2, (a^2 - a + b^2)/(2b))Alright, moving on to Z, the circumcenter of triangle ADE.Points of triangle ADE: A(a, b), D(2,0), E(3,0)Find perpendicular bisectors of AD and AE.Midpoint of AD: ((a + 2)/2, (b + 0)/2) = ((a + 2)/2, b/2)Slope of AD: (b - 0)/(a - 2) = b/(a - 2)Perpendicular bisector slope: -(a - 2)/bEquation: y - b/2 = [-(a - 2)/b](x - (a + 2)/2)Midpoint of AE is ((a + 3)/2, b/2), same as before.Slope of AE is b/(a - 3), so perpendicular bisector slope is -(a - 3)/bEquation: y - b/2 = [-(a - 3)/b](x - (a + 3)/2)Wait, these are the same equations as for X, except for the first one. Wait, no, for Z, we're dealing with triangle ADE, so we need to find the intersection of the perpendicular bisectors of AD and AE.But we already have the perpendicular bisector of AE from before, which is the same as for X. So, let's write both equations.First equation (perpendicular bisector of AD):y - b/2 = [-(a - 2)/b](x - (a + 2)/2)Second equation (perpendicular bisector of AE):y - b/2 = [-(a - 3)/b](x - (a + 3)/2)Set them equal:[-(a - 2)/b](x - (a + 2)/2) = [-(a - 3)/b](x - (a + 3)/2)Multiply both sides by b:-(a - 2)(x - (a + 2)/2) = -(a - 3)(x - (a + 3)/2)Simplify signs:(a - 2)(x - (a + 2)/2) = (a - 3)(x - (a + 3)/2)Expand both sides:Left: (a - 2)x - (a - 2)(a + 2)/2Right: (a - 3)x - (a - 3)(a + 3)/2Simplify:Left: (a - 2)x - (a^2 - 4)/2Right: (a - 3)x - (a^2 - 9)/2Bring all terms to left:(a - 2)x - (a^2 - 4)/2 - (a - 3)x + (a^2 - 9)/2 = 0Factor x terms:[(a - 2) - (a - 3)]x + [ - (a^2 - 4)/2 + (a^2 - 9)/2 ] = 0Simplify:[1]x + [ (-a^2 + 4 + a^2 - 9)/2 ] = 0Which is:x + (-5)/2 = 0 => x = 5/2Substitute x = 5/2 into first equation:y - b/2 = [-(a - 2)/b](5/2 - (a + 2)/2) = [-(a - 2)/b]( (5 - a - 2)/2 ) = [-(a - 2)/b]( (3 - a)/2 )So,y = b/2 + [-(a - 2)(3 - a)]/(2b)Simplify numerator:-(a - 2)(3 - a) = -( -a^2 + 5a - 6 ) = a^2 - 5a + 6So,y = b/2 + (a^2 - 5a + 6)/(2b) = [b^2 + a^2 - 5a + 6]/(2b)Therefore, circumcenter Z is at (5/2, (a^2 - 5a + b^2 + 6)/(2b))Wait, let me check that calculation again. When I substituted x = 5/2, I had:y - b/2 = [-(a - 2)/b]*(5/2 - (a + 2)/2) = [-(a - 2)/b]*( (5 - a - 2)/2 ) = [-(a - 2)/b]*( (3 - a)/2 )So, that's [-(a - 2)(3 - a)]/(2b) = [-( -a^2 + 5a - 6 )]/(2b) = (a^2 - 5a + 6)/(2b)So, y = b/2 + (a^2 - 5a + 6)/(2b) = [b^2 + a^2 - 5a + 6]/(2b)Yes, that's correct.Now, finally, T is the circumcenter of triangle ACD.Points of triangle ACD: A(a, b), C(1,0), D(2,0)Find perpendicular bisectors of AC and AD.Midpoint of AC: ((a + 1)/2, b/2)Slope of AC: (b - 0)/(a - 1) = b/(a - 1)Perpendicular bisector slope: -(a - 1)/bEquation: y - b/2 = [-(a - 1)/b](x - (a + 1)/2)Midpoint of AD: ((a + 2)/2, b/2)Slope of AD: (b - 0)/(a - 2) = b/(a - 2)Perpendicular bisector slope: -(a - 2)/bEquation: y - b/2 = [-(a - 2)/b](x - (a + 2)/2)So, we have two equations:1. y = [-(a - 1)/b](x - (a + 1)/2) + b/22. y = [-(a - 2)/b](x - (a + 2)/2) + b/2Set them equal:[-(a - 1)/b](x - (a + 1)/2) + b/2 = [-(a - 2)/b](x - (a + 2)/2) + b/2Subtract b/2:[-(a - 1)/b](x - (a + 1)/2) = [-(a - 2)/b](x - (a + 2)/2)Multiply both sides by b:-(a - 1)(x - (a + 1)/2) = -(a - 2)(x - (a + 2)/2)Simplify signs:(a - 1)(x - (a + 1)/2) = (a - 2)(x - (a + 2)/2)Expand both sides:Left: (a - 1)x - (a - 1)(a + 1)/2Right: (a - 2)x - (a - 2)(a + 2)/2Simplify:Left: (a - 1)x - (a^2 - 1)/2Right: (a - 2)x - (a^2 - 4)/2Bring all terms to left:(a - 1)x - (a^2 - 1)/2 - (a - 2)x + (a^2 - 4)/2 = 0Factor x terms:[(a - 1) - (a - 2)]x + [ - (a^2 - 1)/2 + (a^2 - 4)/2 ] = 0Simplify:[1]x + [ (-a^2 + 1 + a^2 - 4)/2 ] = 0Which is:x + (-3)/2 = 0 => x = 3/2Substitute x = 3/2 into first equation:y = [-(a - 1)/b](3/2 - (a + 1)/2) + b/2 = [-(a - 1)/b]*( (3 - a - 1)/2 ) + b/2 = [-(a - 1)/b]*( (2 - a)/2 ) + b/2Simplify:y = [-(a - 1)(2 - a)]/(2b) + b/2 = [-( -a^2 + 3a - 2 )]/(2b) + b/2 = (a^2 - 3a + 2)/(2b) + b/2Combine terms:y = (a^2 - 3a + 2 + b^2)/(2b)So, circumcenter T is at (3/2, (a^2 - 3a + b^2 + 2)/(2b))Wait, let me verify that:From the calculation:y = [-(a - 1)(2 - a)]/(2b) + b/2= [-(a - 1)(- (a - 2))]/(2b) + b/2= [(a - 1)(a - 2)]/(2b) + b/2= (a^2 - 3a + 2)/(2b) + b/2Yes, that's correct.So, T is at (3/2, (a^2 - 3a + b^2 + 2)/(2b))Now, we have coordinates for X, Y, Z, and T.Let me summarize:- X: (3/2, (a^2 - 3a + b^2)/(2b))- Y: (1/2, (a^2 - a + b^2)/(2b))- Z: (5/2, (a^2 - 5a + b^2 + 6)/(2b))- T: (3/2, (a^2 - 3a + b^2 + 2)/(2b))Now, I need to show that T is the centroid of triangle XYZ.The centroid of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is given by:G = ( (x1 + x2 + x3)/3, (y1 + y2 + y3)/3 )So, let's compute the centroid G of triangle XYZ.Compute x-coordinate:( X_x + Y_x + Z_x ) / 3 = (3/2 + 1/2 + 5/2)/3 = ( (3 + 1 + 5)/2 ) /3 = (9/2)/3 = 3/2Similarly, y-coordinate:( X_y + Y_y + Z_y ) / 3Compute each y-coordinate:X_y = (a^2 - 3a + b^2)/(2b)Y_y = (a^2 - a + b^2)/(2b)Z_y = (a^2 - 5a + b^2 + 6)/(2b)Sum:[ (a^2 - 3a + b^2) + (a^2 - a + b^2) + (a^2 - 5a + b^2 + 6) ] / (2b)Simplify numerator:a^2 - 3a + b^2 + a^2 - a + b^2 + a^2 - 5a + b^2 + 6Combine like terms:3a^2 - 9a + 3b^2 + 6So, sum is (3a^2 - 9a + 3b^2 + 6)/(2b)Divide by 3:(3a^2 - 9a + 3b^2 + 6)/(6b) = (a^2 - 3a + b^2 + 2)/(2b)Therefore, centroid G is at (3/2, (a^2 - 3a + b^2 + 2)/(2b))But wait, that's exactly the coordinates of T!So, T coincides with the centroid G of triangle XYZ.Therefore, T is the centroid of triangle XYZ.That's the proof.

💡Okay, so I've got this problem about two runners starting from points A and B, which are 5 miles apart. They run towards each other and meet after 1 hour. Then, if they run in the same direction, starting from A and B respectively, the faster one catches up in 3 hours. I need to find the ratio of the speed of the faster runner to the slower runner. The options are A) 1.5, B) 2, C) 2.5, and D) 3.Alright, let's break this down. First, when they run towards each other, they meet after 1 hour. So, the combined distance they cover in that hour is 5 miles. That means their speeds add up to 5 miles per hour. Let's call the speed of the faster runner u1 and the slower one u2. So, u1 + u2 = 5.Now, when they run in the same direction, starting from A and B, the faster runner catches up in 3 hours. Since they're moving in the same direction, the relative speed between them is u1 - u2. The distance between them is still 5 miles, so the time it takes for the faster runner to catch up is the distance divided by the relative speed. That gives us 5 = (u1 - u2) * 3. So, u1 - u2 = 5/3.Now we have two equations:1. u1 + u2 = 52. u1 - u2 = 5/3I can solve these two equations to find u1 and u2. Let's add the two equations together. Adding the left sides: u1 + u2 + u1 - u2 = 2u1. Adding the right sides: 5 + 5/3 = 15/3 + 5/3 = 20/3. So, 2u1 = 20/3, which means u1 = 10/3.Now, subtract the second equation from the first. Subtracting the left sides: u1 + u2 - (u1 - u2) = 2u2. Subtracting the right sides: 5 - 5/3 = 15/3 - 5/3 = 10/3. So, 2u2 = 10/3, which means u2 = 5/3.Now, to find the ratio of the faster runner's speed to the slower runner's speed, I'll divide u1 by u2. That is (10/3) / (5/3) = (10/3) * (3/5) = 10/5 = 2.So, the ratio is 2. That corresponds to option B.Wait, let me double-check my calculations to make sure I didn't make a mistake. When they run towards each other, their combined speed is 5 mph, which seems right because they cover 5 miles in 1 hour. When running in the same direction, the relative speed is 5/3 mph, which means the faster runner is indeed 2 times faster than the slower one. Yeah, that makes sense. So, I think I did it correctly.

💡First, I need to determine the total number of red cards in the deck. There are 3 red suits, each containing 12 cards, so there are 3 × 12 = 36 red cards out of 60 total cards.Next, I calculate the probability of drawing a red card on the first draw, which is 36/60.After drawing one red card, there are now 35 red cards left out of a total of 59 remaining cards. Therefore, the probability of drawing a second red card is 35/59.Finally, to find the probability that both the first and second cards drawn are red, I multiply the probabilities of each individual event: (36/60) × (35/59) = 21/59.

💡To determine the value of ( a ) for which the quadratic equation ( 2x^2 - ax + 2 = 0 ) has two equal real roots, we can use the discriminant condition for a quadratic equation. 1. **Identify the coefficients**: The given quadratic equation is ( 2x^2 - ax + 2 = 0 ). Here, ( A = 2 ), ( B = -a ), and ( C = 2 ).2. **Discriminant Condition**: For a quadratic equation ( Ax^2 + Bx + C = 0 ) to have two equal real roots, the discriminant must be zero. The discriminant ( Delta ) is given by: [ Delta = B^2 - 4AC ] 3. **Set the Discriminant to Zero**: [ (-a)^2 - 4 times 2 times 2 = 0 ] 4. **Simplify the Equation**: [ a^2 - 16 = 0 ] 5. **Solve for ( a )**: [ a^2 = 16 implies a = pm 4 ] 6. **Conclusion**: The value of ( a ) is ( pm 4 ), which corresponds to option B.