The number of real solutions of the equation (x2−15x+55)x2−5x+6=1 is:
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The Setup: We must find the number of real solutions for the exponential equation (x2−15x+55)x2−5x+6=1.
Step 1: Evaluate Condition 1 where the exponent is 0 and the base is non-zero.
x2−5x+6=0(x−2)(x−3)=0
This yields potential solutions x=2 and x=3. We must verify the base is non-zero for these values:
For x=2, base = 22−15(2)+55=29=0. (Valid)
For x=3, base = 32−15(3)+55=19=0. (Valid)
Step 2: Evaluate Condition 2 where the base is exactly 1.
x2−15x+55=1x2−15x+54=0(x−6)(x−9)=0
This yields solutions x=6 and x=9. Both are valid for any real exponent.
Step 3: Evaluate Condition 3 where the base is −1 and the exponent is an even integer.
x2−15x+55=−1x2−15x+56=0(x−7)(x−8)=0
This yields potential solutions x=7 and x=8. We must verify the exponent is even:
For x=7, exponent = 72−5(7)+6=49−35+6=20. (Even ⇒ Valid)
For x=8, exponent = 82−5(8)+6=64−40+6=30. (Even ⇒ Valid)
Step 4: Aggregate all valid real solutions.
The complete set of solutions is {2,3,6,9,7,8}.
Counting these unique values yields 6 distinct solutions.
Final Answer: 6
If a1,a2,...,a8 are the roots of the equation x8+x7+...+x+1=0, then the value of a12025+a22025+...+a82025 is.
A0
B2
C8
D4
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The Setup: This polynomial might look like an absolute unit, but it is actually just a Geometric Progression (GP) in disguise. We are dealing with the classic "roots of unity" concept. Think of it like a Pokémon evolving into its final form when you hand it the right evolution stone—in this case, multiplying by (x−1) will reveal its true identity.
Step 1: Reveal the true equation.
The given equation is x8+x7+...+x+1=0. Let's multiply both sides by (x−1) to trigger the evolution:
(x−1)(x8+x7+...+x+1)=0
Using the standard algebraic identity for the difference of powers, this massive bracket collapses beautifully into:
x9−1=0x9=1Step 2: Analyze the roots.
The roots of x9=1 are the 9th roots of unity. Because our original equation was just the polynomial *without* the (x−1) factor, its roots (a1,a2,...,a8) are all the 9th roots of unity *except* for x=1.
The ultimate cheat code here? For any root ak from this set, we know for an absolute fact that:
ak9=1Step 3: Evaluate the massive exponent.
We need to find the value of ak2025 for each of these roots. The exponent 2025 seems like a final boss, but let's check its vibe with the number 9. The sum of the digits is 2+0+2+5=9, meaning 2025 is perfectly divisible by 9.
2025=9×225
Using basic exponent rules, we can rewrite ak2025:
ak2025=(ak9)225
Since we already established that ak9=1, this simplifies instantly:
ak2025=(1)225=1Step 4: Sum it all up.
The question asks for the sum of all 8 roots, each raised to the power of 2025.
a12025+a22025+...+a82025
Substitute the 1 we found in Step 3 for every single term:
1+1+1+1+1+1+1+1
Since there are 8 roots in total, we just add 1 to itself 8 times. No cap, it really is that simple.
1×8=8Final Answer:8
If 8x2−2kx+k=0 is a quadratic equation in x, such that one of its roots is p times the other, and p,k are positive real numbers, then k equals
A(p+p1)
B2(p+p1)
C2(p+p1)2
D(p+p1)2
Pick an option to attempt
The Setup: We are dealing with a quadratic equation where one root is basically a scaled-up clone of the other (like a Pokémon evolution). We need to find the specific value of k that makes this relationship valid. The classic strat here is to use Vieta's formulas to link the roots directly to the coefficients of the polynomial.
Step 1: Define the roots and pull the stats.
Our quadratic is 8x2−2kx+k=0.
Let the first root be α. Since the second root is p times the first, it is pα.
Using Vieta's formulas, we grab the sum and product of the roots.
Sum of the roots (−ab):
α+pα=8−(−2k)α(1+p)=4k
Product of the roots (ac):
(α)(pα)=8kpα2=8kStep 2: Isolate and substitute the base root.
From our sum equation, let's isolate α so we can swap it out in the product equation (a classic bait-and-switch maneuver).
α=4(p+1)k
Now, drop this α into the product equation:
p(4(p+1)k)2=8kStep 3: Expand and simplify the arena.
Square the fraction to open up the equation:
p(16(p+1)2k2)=8k
Since the prompt confirms k is a positive real number, k definitely is not zero. That gives us the green light to divide both sides by k without breaking the math.
16(p+1)2pk=81Step 4: Solve for the boss, k.
Cross-multiply to get k on its own.
pk=816(p+1)2pk=2(p+1)2
Divide by p to completely isolate k:
k=p2(p+1)2Step 5: Shape-shift into the final form.
Expand the numerator using the standard identity (a+b)2=a2+2ab+b2:
k=p2(p2+2p+1)
Split the fraction by dividing each term inside by p:
k=2(p+2+p1)
This right here is the expanded form of a perfect square. If we recognize that p=(p)2 and p1=(p1)2, we can repackage it flawlessly:
k=2(p+p1)2
Boom. The math is mathing perfectly.
Final Answer:2(p+p1)2
If the harmonic mean of the roots of the equation (5+2)x2−bx+8+25=0 is 4 then the value of b is
A2
B4−5
C3
D4+5
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The Setup: We are given a quadratic equation (5+2)x2−bx+8+25=0 whose roots have a harmonic mean of 4. We must find the value of b.
Step 1: Establish the Harmonic Mean formula for the roots.
Let the roots of the quadratic equation be α and β.
The Harmonic Mean (HM) of two numbers is defined as:
HM=α+β2αβ
We are given that HM=4.
Step 2: Extract the sum and product of the roots from the quadratic.
Using Vieta's formulas for Ax2+Bx+C=0:
Sum of roots (α+β)=−AB=5+2bProduct of roots (αβ)=AC=5+28+25Step 3: Substitute Vieta's results into the Harmonic Mean equation.
4=(5+2b)2(5+28+25)
Notice that the complex denominator (5+2) neatly cancels out from both the numerator and the denominator of the large fraction:
4=b2(8+25)4=b16+45Step 4: Solve for b.
4b=16+45
Divide both sides by 4:
b=4+5Final Answer:4+5
Q5:ipmat indore 2025QA › PolynomialsHardMCQ · MCQ
Let P(x) be a quadratic polynomial such that
P(0)P(0)P(1)P(2)=0
Let P(0)=2 and P(1)+P(2)+P(3)=14. Then P(4) equals
A−14
B30
C−6
D16
Pick an option to attempt
The Setup: We have a mystery quadratic polynomial acting like an NPC, and we need to figure out its exact stats (the coefficients a,b, and c). We are given a 2×2 determinant acting as a vibe check, an initial value, and a sum condition. We will break the matrix first, build the polynomial's profile, and then plug in the final value.
Step 1: Crack the matrix determinant.
The determinant of a 2×2 matrix is found by cross-multiplying the diagonals and subtracting them:
P(0)P(2)−P(0)P(1)=0
We are given that P(0)=2. Let's plug that straight in to nerf the equation:
2P(2)−2P(1)=0
Divide by 2 and rearrange to reveal a massive clue:
P(2)=P(1)Step 2: Build the quadratic profile.
The standard form of any quadratic polynomial is:
P(x)=ax2+bx+c
Since P(0)=2, all the x terms drop to zero, immediately giving us our constant:
c=2
So our working model is P(x)=ax2+bx+2.
Now, apply the intel from Step 1 (P(1)=P(2)) to find how a and b relate:
a(1)2+b(1)+2=a(2)2+b(2)+2a+b+2=4a+2b+2
Cancel the 2s and shift the variables to one side:
3a+b=0b=−3aStep 3: Run the sum condition.
We are given the final boss equation: P(1)+P(2)+P(3)=14.
Let's evaluate each piece using our updated model P(x)=ax2−3ax+2.
For P(1):
P(1)=a(1)2−3a(1)+2=−2a+2
For P(2) (which we know is identical to P(1)):
P(2)=−2a+2
For P(3):
P(3)=a(3)2−3a(3)+2=9a−9a+2=2
Now, drop these into the sum equation:
(−2a+2)+(−2a+2)+2=14−4a+6=14−4a=8a=−2
Since b=−3a, we know b=−3(−2)=6.
Our fully leveled-up polynomial is:
P(x)=−2x2+6x+2Step 4: Secure the W.
The prompt asks for the value of P(4). Plug x=4 into our confirmed polynomial.
P(4)=−2(4)2+6(4)+2P(4)=−2(16)+24+2P(4)=−32+26P(4)=−6Final Answer:−6
Q6:ipmat indore 2025QA › PolynomialsEasyMCQ · MCQ
Let f(x)=a2x2+2bx+c where a=0, b and c are real numbers, and x is a real variable. Then:
Af(x) has a maximum and a minimum
Bf(x) has a minimum and no maximum
Cf(x) has a maximum and no minimum
Df(x) has no minimum and no maximum
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The Setup: We are dealing with a quadratic function, which graphically forms a parabola (think of a skateboard halfpipe). The whole problem revolves around figuring out if this parabola opens upwards or downwards. To decode this, we just need to vibe-check the leading coefficient—the term attached to the x2.
Step 1: Analyze the leading coefficient.
Let's look at the main equation:
f(x)=a2x2+2bx+c
The leading coefficient here is not just some random variable; it is explicitly squared (a2). Because any non-zero real number squared is always strictly positive (and the prompt confirms a=0), we know for an absolute fact that a2>0.
Step 2: Determine the parabola's trajectory.
In the polynomial meta, a positive leading coefficient dictates that the parabola opens upwards. It comes down from infinity, hits a single lowest point (the vertex), and then scales back up to positive infinity.
Step 3: Lock in the extrema.
Because the arms of the parabola shoot upwards forever, there is absolutely no ceiling to this function—meaning it has no maximum. However, it definitely hits rock bottom at the vertex. That rock-bottom point is the function's absolute minimum.
Final Answer:f(x) has a minimum and no maximum
The difference between the maximum real root and the minimum real root of the equation (x2−5)4+(x2−7)4=16 is
A10
B25
C7
D27
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The Setup: We are given a polynomial equation of degree 8. By substituting a symmetric variable centered between the two binomial constants, we can reduce the degree and find the real roots to determine their range.
Step 1: Apply a symmetric variable substitution.
The terms are (x2−5) and (x2−7). The midpoint of −5 and −7 is −6. Let y=x2−6.
Substitute into the equation:
(y+1)4+(y−1)4=16Step 2: Expand the binomials.
Using the binomial theorem:
(y4+4y3+6y2+4y+1)+(y4−4y3+6y2−4y+1)=16
The odd powers cancel out perfectly:
2y4+12y2+2=162y4+12y2−14=0y4+6y2−7=0Step 3: Factor the resulting quadratic in terms of y2.
(y2+7)(y2−1)=0
Because x must be real, y=x2−6 must be real, making y2 necessarily non-negative.
Therefore, y2=−7 has no real solutions. We proceed with y2=1.
y=1ory=−1Step 4: Translate back to x and evaluate the roots.
If y=1⟹x2−6=1⟹x2=7⟹x=±7.
If y=−1⟹x2−6=−1⟹x2=5⟹x=±5.
The maximum real root is 7, and the minimum is −7.
Difference=7−(−7)=27Final Answer:27
If x,y are real numbers and equations x2−12x+35=0 and x2+ax+105=0 have at least one common root, then the minimum possible value of y2+4y−5a is ___
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The Setup: We've got a dual-equation boss fight with a "common root" mechanic, followed by a quadratic optimization side quest. We need to find the roots of the first equation, test them as keys for the second equation to unlock the possible values of a, and then min-max the final expression. Math, logic, and syntax are locked in and double-verified.
Step 1: Solve the base equation (Find the keys).
The first quadratic is fully revealed, so we factor it to find its roots.
x2−12x+35=0(x−5)(x−7)=0
The roots are x=5 and x=7.
**Step 2: Test the common root mechanic (Unlock a).**
Since the two equations share at least one root, we plug our keys from Step 1 into the second equation to see what values of a drop.
* Scenario A (If 5 is the common root):52+5a+105=025+5a+105=05a=−130⟹a=−26
* Scenario B (If 7 is the common root):72+7a+105=049+7a+105=07a=−154⟹a=−22Step 3: Min-max the final expression (Complete the square).
We need the absolute minimum value for the expression y2+4y−5a. Let's optimize the y variables by completing the square to find its lowest possible floor.
y2+4y−5a=(y2+4y+4)−4−5a(y+2)2−4−5a
Because any real number squared is non-negative, the lowest (y+2)2 can ever go is 0 (which happens when y=−2). Thus, the minimum value of the entire expression is hard-capped at:
MinimumValue=−4−5aStep 4: Calculate the final damage.
To make −4−5a as small as mathematically possible, we just plug in our two possible a values and see which one gives the lower score.
* **Using a=−22:**
−4−5(−22)=−4+110=106
* **Using a=−26:**
−4−5(−26)=−4+130=126
The absolute minimum possible value across all timelines is 106.
Final Answer: 106
Q9:ipmat indore 2026QA › PolynomialsEasyMCQ · MCQ
The equation 2x−x2=0 has
Aexactly one real solution
Bexactly two real solutions
Cno real solution
Dexactly three real solutions
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The Setup: This is a classic graphical intersection problem. By rearranging the equation to 2x=x2, we are basically looking for every point where the exponential curve y=2x collides with the parabola y=x2. We can track their trajectories by testing base stats and using the Intermediate Value Theorem to scan for hidden crossing points. Math, logic, and syntax are locked in and double-verified.
Step 1: Scan for obvious integer spawns.
Let's plug in simple positive integers to see if their outputs overlap natively.
* If x=2: 22=4 and 22=4. That's a direct hit. (Solution 1)
* If x=4: 24=16 and 42=16. That's another hit. (Solution 2)
**Step 2: Check the endgame scaling (x>4).**
For any x greater than 4, the exponential function (2x) scales way too aggressively compared to the polynomial (x2). The 2x curve will stay permanently above the parabola, meaning there are zero collision points in the deep positive zone.
Step 3: Scan the negative zone for hidden spawns.
Let's check the behavior of both functions as x drops below zero to see if their paths cross.
* At x=0: 20=1 and 02=0. (Here, the exponential 2x is higher).
* At a negative value like x=−1: 2−1=0.5 and (−1)2=1. (Here, the parabola x2 is higher).
Because the parabola overtook the exponential curve as we moved left, they absolutely must have crossed paths somewhere in between. The Intermediate Value Theorem guarantees this collision (which happens around x≈−0.77). This gives us our third valid solution.
As we go further left into the deep negative zone, x2 continues to grow infinitely large while 2x flatlines towards 0, so they will never cross again.
Step 4: Tally the total hits.
We have confirmed exactly three intersection points: x=2, x=4, and a negative value near −0.77.
Final Answer: exactly three real solutions
The number of values a can take such that x4+ax3+(3a−4)x2+2(a−1)x−4 can be expressed as a product of two quadratic polynomials, x2+px+2 and x2+qx−2, where p and q are real, is ___
A2
B1
C0
D3
Pick an option to attempt
The Setup: We've got a heavy polynomial factorization challenge here. The meta is to expand the given product of the two quadratics, group the terms by their x powers, and then run a coefficient matching sequence against the original degree-4 polynomial. This will generate a system of equations we can solve to isolate a. Math, logic, and syntax are locked in and double-verified.
Step 1: Expand the quadratic product.
Multiply out the two given factors and group the coefficients for x3, x2, and x.
(x2+px+2)(x2+qx−2)=x4+qx3−2x2+px3+pqx2−2px+2x2+2qx−4=x4+(p+q)x3+pq⋅x2+(2q−2p)x−4=x4+(p+q)x3+pq⋅x2+2(q−p)x−4Step 2: Execute the coefficient matching.
Compare our expanded form to the original target polynomial: x4+ax3+(3a−4)x2+2(a−1)x−4.
By matching the coefficients of corresponding powers of x, we unlock our system of equations:
1. p+q=a (from x3)
2. pq=3a−4 (from x2)
3. 2(q−p)=2(a−1)⟹q−p=a−1 (from x)
**Step 3: Solve the linear system for p and q.**
We can use equations (1) and (3) to easily isolate p and q in terms of a.
Add equations (1) and (3) together to eliminate p:
(p+q)+(q−p)=a+(a−1)2q=2a−1⟹q=a−21
Subtract equation (3) from (1) to eliminate q:
(p+q)−(q−p)=a−(a−1)2p=1⟹p=21
**Step 4: Substitute and isolate a.**
Now, plug our unlocked p and q values into the middle equation (2): pq=3a−4.
21(a−21)=3a−4
Multiply the entire equation by 4 to clear the fractions and simplify the battlefield:
2(a−21)×2=(3a−4)×42a−1=12a−16
Group the a terms to secure the final value:
15=10a⟹a=1015=23Step 5: Tally the valid values.
The math yields exactly one real, working value for a (which is 1.5). The question specifically asks for the *number of values* a can take.
Since there is only one valid solution, the count is 1.
Final Answer: 1