Q.If ax2+bx+c is divided by x+3, x−5, and x−1, the remainders are 21, 61 and 9 respectively. Find a,b and c. (Use Gaussian elimination method.)
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Concept understanding — Gaussian Elimination
Gaussian elimination solves AX=B by row-reducing the augmented matrix[A∣B] to row-echelon form using elementary row operations, then reading off the unknowns from the bottom equation upward -- the method of back substitution. Unlike matrix inversion or Cramer's rule, it works even when the coefficient matrix is singular or non-square (any number of equations, any number of unknowns), which is exactly why it is the most general of the three methods.
Worked illustration. For x+y=3,2x−y=0: [A∣B]=(121−1∣∣30)R2→R2−2R1(101−3∣∣3−6). The second row reads −3y=−6⇒y=2; back-substituting into the first row, x+2=3⇒x=1 -- matching the Cramer's-rule answer above.
The general shape for a 3×3 system. Row-reduce [A∣B] to 100∗10∗∗1∣∣∣∗∗∗-type echelon form; the bottom row gives z directly, the middle row (with z substituted) gives y, and the top row (with y,z substituted) gives x.
Word problems -- a quadratic/cubic passing through given points, a polynomial's remainders under the Remainder Theorem, money split across interest-bearing bonds, a projectile's path -- all translate into a linear system whose augmented matrix is then row-reduced exactly as above; the method is identical whether the unknowns are prices, rates, or polynomial coefficients.
Balancing a chemical equation. Writing a reaction x1(reactant1)+x2(reactant2)→x3(product1)+x4(product2) and demanding the atom count of each element balance on both sides gives one homogeneous linear equation per element in the unknowns x1,…,x4. Gaussian elimination on this homogeneous system typically leaves one unknown free; choosing the smallest value that makes every xi a positive integer gives the balanced equation.
Tip
Because Gaussian elimination never needs A to be square, it is the natural tool whenever a word problem produces more equations than unknowns (an over-determined system) or fewer (an under-determined one) -- situations where matrix inversion and Cramer's rule simply do not apply.
Use the remainder theorem to turn each division fact into a linear equation in a,b,c, then solve the 3×3 system by Gaussian elimination.
f(−3)=21,f(5)=61,f(1)=9 give a=2,b=1,c=6.
✓Final answer
a=2,b=1,c=6, i.e. f(x)=2x2+x+6.
By the remainder theorem, dividing f(x)=ax2+bx+c by x−k leaves remainder f(k); the three given remainders give three linear equations in a,b,c, which we solve by Gaussian elimination.
Step 1. Translate the remainders into equations. Dividing by x+3=x−(−3) leaves remainder f(−3)=21; dividing by x−5 leaves f(5)=61; dividing by x−1 leaves f(1)=9.
f(−3)=9a−3b+c=21,f(5)=25a+5b+c=61,f(1)=a+b+c=9
Step 2. Reorder so the simplest (unit-coefficient) equation leads -- equivalent to a row swap.
19251−3511192161
Step 3. Eliminate a from R2,R3 using R1. Apply R2→R2−9R1 and R3→R3−25R1:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2023Set ANNUAL5 marks
Q.(a) A boy is walking along the path y=ax2+bx+c through the points (−6,8), (−2,−12) and (3,8). He wants to meet his friend at P(7,60). Will he meet his friend ? (Use Gaussian Elimination method)
OR
(b) Prove that the ellipse x2+4y2=8 and the hyperbola x2−2y2=4 intersect orthogonally.
›Reveal solutionSolution
(a) Sets up and Gaussian-eliminates a 3-equation linear system for a,b,c in y=ax2+bx+c, then checks whether (7,60) lies on the resulting parabola; (b) differentiates both conics implicitly and shows the product of their slopes is −1 at every intersection point. Both alternatives answered below.
(a) Path through 3 points — Gaussian elimination
1. Set up equations. Substituting each point into y=ax2+bx+c:
3. Solve the reduced 2×2 system. From (i): b=8a−5. Substitute into (ii): a+(8a−5)=4⇒9a=9⇒a=1. Then b=8(1)−5=3.
4. Back-substitute for c. Using Eq.2: 4(1)−2(3)+c=−12⇒4−6+c=−12⇒c=−10.
5. Path equation.y=x2+3x−10. Check against all three original points: (−6,8): 36−18−10=8✓; (−2,−12): 4−6−10=−12✓; (3,8): 9+9−10=8✓.
6. Test the friend's point (7,60).y(7)=72+3(7)−10=49+21−10=60 — exactly matches. Yes, the boy will meet his friend at P(7,60), since (7,60) lies on his path.
(b) Orthogonal intersection of x2+4y2=8 and x2−2y2=4
1. Find the points of intersection. From the hyperbola, x2=4+2y2. Substitute into the ellipse: (4+2y2)+4y2=8⇒6y2=4⇒y2=32. Then x2=4+2(32)=4+34=316.
2. Slopes on each curve. Differentiating the ellipse x2+4y2=8 implicitly: 2x+8yy′=0⇒m1=dxdy=−4yx.
Differentiating the hyperbola x2−2y2=4 implicitly: 2x−4yy′=0⇒m2=dxdy=2yx.
3. Product of slopes at the intersection.
m1m2=(−4yx)(2yx)=−8y2x2
4. Substitute the intersection valuesx2=316, y2=32:
m1m2=−8(2/3)16/3=−16/316/3=−1
5. Conclusion. Since m1m2=−1 at every point of intersection, the tangents to the two curves are perpendicular there — the ellipse and hyperbola intersect orthogonally.