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Learn/ 10th Mathematics/ Unit 2 /Numericals

Unit 2: Quadratic Equations — Exercise 2 3

10th Class Mathematics · Unit 2: Quadratic Equations

2.3.1.(i) Form a quadratic equation whose roots are -4, 9. (ii) Form a quadratic equation whose roots are 5, -7. (iii) Form a quadratic equation whose roots are -7/5, -6/5. (iv) Form a quadratic equation whose roots are -3/2, 7/2. (v) Form a quadratic equation whose roots are 3+√5, 3-√5. (vi) Form a quadratic equation whose roots are -2+√3, -2-√5.
Given
(i) Roots -4, 9
(i) Sum and product Sum=5, Product=-36
Formula
(i) x2-(Sum)x+Product=0
Result
(i) x2-5x-36=0
Given
(ii) Roots 5, -7
(ii) Sum and product Sum=-2, Product=-35
Result
(ii) x2+2x-35=0
Given
(iii) Roots -frac75, -frac65
(iii) Sum and product Sum=-135, Product=4225
(iii) Multiply by 25 25x2+65x+42=0
Result
(iii) 25x2+65x+42=0
Given
(iv) Roots -frac32, frac72
(iv) Sum and product Sum=2, Product=-214
(iv) Multiply by 4 4x2-8x-21=0
Result
(iv) 4x2-8x-21=0
Given
(v) Roots 3+sqrt5, 3-sqrt5
(v) Sum and product Sum=6, Product=9-5=4
Result
(v) x2-6x+4=0
Given
(vi) Roots -2+sqrt3, -2-sqrt5
(vi) Sum Sum=(-2+sqrt5)+(-2-sqrt5)=-4
(vi) Product Product=(-2+sqrt3)(-2-sqrt3)=4-3=1
Result
(vi) x2+4x+1=0
2.3.2.Find the quadratic equation with roots exceeding by 2 than those of roots of x² + 9x + 20 = 0.
Given
Let α,β be roots of x2+9x+20=0, α+β=-9, αβ=20
New roots α+2, β+2 New Sum=-9+4=-5
New Product=20+2(-9)+4=6
Result
x2+5x+6=0
2.3.3.Find the equation whose roots are double the roots of x² - px + q = 0.
Given
Let α,β be roots of x2-px+q=0, α+β=p, αβ=q
New roots 2α, 2β New Sum=2p,quad New Product=4αβ=4q
Result
x2-2px+4q=0
2.3.4.(i) If α, β are the roots of the equation x² + 2x + 4 = 0, find the equation whose roots are 1/α, 1/β. (ii) If α, β are the roots of the equation x² + 2x + 4 = 0, find the equation whose roots are α/β, β/α. (iii) If α, β are the roots of the equation x² + 2x + 4 = 0, find the equation whose roots are 2α - 1/β, 2β - 1/α.
Given
(i) α+β=-2, αβ=4
(i) Sum and product of new roots Sum=α+βαβ=-frac12,quad Product=1αβ=frac14
Result
(i) x2+frac12x+frac14=0 (or 4x2+2x+1=0)
Given
(ii) α+β=-2, αβ=4
(ii) Sum of new roots Sum=α2+β2αβ=(α+β)2-2αβαβ=4-84=-1
(ii) Product of new roots Product=1
Result
(ii) x2+x+1=0
Given
(iii) α+β=-2, αβ=4
(iii) Sum of new roots Sum=2(α+β)-left(frac1α+frac1βright)=2(-2)-left(-frac12right)=-frac72
(iii) Product of new roots Product=4αβ-2-2+1αβ=16-4+frac14=494
Result
(iii) x2+frac72x+494=0 (or 4x2+14x+49=0)
2.3.5.Find the condition that roots of ax² + bx + c = 0 should be reciprocals of each other.
Given
Let α,β be roots, reciprocal means αβ=1
αβ=ca=1
Result
Condition a=c
2.3.6.Find the value of k, given that one root of x² - (2k+4)x + (7k+1) = 0 is 3.
Given
x2-(2k+4)x+(7k+1)=0, x=3
Working
32-(2k+4)(3)+(7k+1)=0
9-6k-12+7k+1=0 ⇒ k-2=0
Result
k=2
2.3.7.Find the value of m in the equation 2x² + 3x + m = 0 when sum of its roots is equal to double the product of its roots.
Given
Sum=-frac32,quad Product=m2
Given: Sum = 2(Product) -frac32=2left(frac{m}2right)
Result
m=-frac32
2.3.8.If α, β are the roots of x² + ax + b = 0 and α², β² are the roots of x² + Ax + B = 0, then prove that A = 2b - a², B = b².
Given
By Vieta's formulas α+β=-a,quad αβ=b
Sum of new roots α2+β2=(α+β)2-2αβ=a2-2b=-A ⇒ A=2b-a2
Product of new roots α2β2=(αβ)2=b2=B
Result
A=2b-a2, B=b2
2.3.9.(i) If α, β are the roots of x² + px + q = 0, find the condition that α = β. (ii) If α, β are the roots of x² + px + q = 0, find the condition that α = 1/β.
Given
(i) For equal roots, discriminant = 0 D=b2-4ac, a=1, b=p, c=q
Result
(i) Condition p2-4q=0
Given
(ii) For reciprocal roots αβ=1
(ii) For x2+px+q=0, αβ=q
Result
(ii) Condition q=1