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Learn/ 9th Mathematics/ Unit 4 /Numericals

Unit 4: Factorization and Algebraic Manipulation — Exercise 4 4

9th Class Mathematics · Unit 4: Factorization and Algebraic Manipulation

4.4.1.Find the square root of the following polynomials by factorization method: (i) x^2 - 8x + 16 (ii) 9x^2 + 12x + 4 (iii) 36a^2 + 84a + 49 (iv) 64y^2 - 32y + 4 (v) 200t^2 - 120t + 18 (vi) 40x^2 + 120x + 90
(i) x2-8x+16 = (x)2-2(x)(4)+(4)2 = (x-4)2
Result
(i) sqrt{x2-8x+16} = ± (x-4)
(ii) 9x2+12x+4 = (3x)2+2(3x)(2)+(2)2 = (3x+2)2
(ii) sqrt{9x2+12x+4} = ± (3x+2)
(iii) 36a2+84a+49 = (6a)2+2(6a)(7)+(7)2 = (6a+7)2
(iii) sqrt{36a2+84a+49} = ± (6a+7)
(iv) 64y2-32y+4 = (8y)2-2(8y)(2)+(2)2 = (8y-2)2
(iv) sqrt{64y2-32y+4} = ± (8y-2)
(v) 200t2-120t+18 = 2[100t2-60t+9] = 2[(10t)2-2(10t)(3)+(3)2] = 2(10t-3)2
The book's next line mistakenly re-labels this result with the previous part's variables, printing '\sqrt{64y^2-32y+18} = \pm\sqrt2(10t-3)' instead of '\sqrt{200t^2-120t+18} = \pm\sqrt2(10t-3)'. The numeric working and final answer are unaffected by this copy-paste label error.
(v) sqrt{200t2-120t+18} = ± sqrt{2}(10t-3)
(vi) 40x2+120x+90 = 10[4x2+12x+9] = 10[(2x)2+2(2x)(3)+(3)2] = 10(2x+3)2
(vi) sqrt{40x2+120x+90} = ± sqrt{10}(2x+3)
4.4.2.Find the square root of the following polynomials by division method: (i) 4x^4 - 28x^3 + 37x^2 + 42x + 9 (ii) 121x^4 - 198x^3 - 183x^2 + 216x + 144 (iii) x^4 - 10x^3y + 27x^2y^2 - 10xy^3 + y^4 (iv) 4x^4 - 12x^3 + 37x^2 - 42x + 49
(i) Division of 4x4-28x3+37x2+42x+9 gives quotient 2x2-7x-3, remainder 0
Result
(i) sqrt{4x4-28x3+37x2+42x+9} = ± (2x2-7x-3)
(ii) Division of 121x4-198x3-183x2+216x+144 gives quotient 11x2-9x-12, remainder 0
(ii) sqrt{121x4-198x3-183x2+216x+144} = ± (11x2-9x-12)
(iii) Division of x4-10x3y+27x2y2-10xy3+y4 gives quotient x2-5xy+y2, remainder 0
(iii) sqrt{x4-10x3y+27x2y2-10xy3+y4} = ± (x2-5xy+y2)
(iv) Division of 4x4-12x3+37x2-42x+49 gives quotient 2x2-3x+7, remainder 0
(iv) sqrt{4x4-12x3+37x2-42x+49} = ± (2x2-3x+7)
4.4.3.An investor's return R(x) in rupees after investing x thousand rupees is given by quadratic expression: R(x) = -x^2 + 6x - 8. Factorize the expression and find the investment levels that result in zero return.
Given
Given R(x) = -x2+6x-8
Factorizing R(x) = -x2+4x+2x-8 = -x(x-4)+2(x-4) = (-x+2)(x-4)
Setting to zero (-x+2)(x-4)=0 implies -x+2=0 or x-4=0
Result
Result x=2 and x=4
4.4.4.A company's profit P(x) in rupees from selling x units of a product is modeled by the cubic expression: P(x) = x^3 - 15x^2 + 75x - 125. Find the break-even point(s), where the profit is zero.
Given
Given P(x) = x3-15x2+75x-125
Factorizing P(x) = (x)3-3(x)2(5)+3(x)(5)2-(5)3 = (x-5)3
Setting to zero (x-5)3=0
Result
Result x=5
4.4.5.The potential energy V(x) in an electric field varies as a cubic function of distance x, given by: V(x) = 2x^3 - 6x^2 + 4x. Determine where the potential energy is zero.
Given
Given V(x) = 2x3-6x2+4x
Factorizing V(x) = 2x(x2-3x+2) = 2x(x-2)(x-1)
Setting to zero 2x(x-1)(x-2)=0
Result
Result x=0, x=1, x=2
4.4.6.In structural engineering, the deflection Y(x) of a beam is given by: Y(x) = 2x^2 - 8x + 6. This equation gives the vertical deflection at any point x along the beam. Find the points of zero deflection.
Given
Given Y(x) = 2x2-8x+6
Factorizing Y(x) = 2x2-8x2+6
This restated line is printed with a typo as '2x^2-8x^2+6' (should repeat the stem's '2x^2-8x+6'); the middle term should be -8x, not -8x^2. All subsequent working correctly uses -8x, so the final result is unaffected.
Factorizing Y(x) = 2(x2-4x+3) = 2(x2-3x-x+3) = 2(x-1)(x-3)
Setting to zero 2(x-1)(x-3)=0 implies 2neq0 so x-1=0 or x-3=0
Result
Result x=1, x=3