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CAT 2025 Lesson : Quadratic Equations - Changes to Roots

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3.5 Changes to Roots

Where α\alpha and β\beta are the roots of a quadratic equation ax2+bx+c=0ax^{2} + bx + c = 0, and kk is a constant,

For the roots The quadratic equation is
(α+k)(\alpha + k) and (β+k)(\beta + k) a(x−k)2+b(x−k)+c=0a(x - k)^{2} + b(x - k) + c = 0
(α−k)(\alpha - k) and (β−k)(\beta - k) a(x+k)2+b(x+k)+c=0a(x + k)^{2} + b(x + k) + c = 0
(αk)(\alpha k) and (βk)(\beta k) a(xk)2+b(xk)+c=0a \left(\dfrac{x}{k} \right)^{2} + b \left(\dfrac{x}{k} \right) + c = 0
(αk)\left(\dfrac{\alpha}{k} \right) and (βk)\left(\dfrac{\beta}{k} \right) a(xk)2+b(xk)+c=0a(xk)^{2} + b(xk) + c = 0
(1α)\left(\dfrac{1}{\alpha} \right) and (1β)\left(\dfrac{1}{\beta} \right) cx2+bx+a=0cx^{2} + bx + a = 0


Ratio of Roots: For a quadratic equation ax2+bx+c=0ax^{2} + bx + c = 0, if the ratio of roots, i.e., α:β=p:q\alpha : \beta = p : q, then b2pq=ac(p+q)2\bm{b^{2}pq = ac(p + q)^{2}}

Example 9

If α\alpha and β\beta are the roots of the quadratic equation 2x2+3x−4=02x^{2} + 3x - 4 = 0, what is the equation whose roots are
(I) α+52\dfrac{\alpha + 5}{2} and β+52\dfrac{\beta + 5}{2} ?
(II) two-third the reciprocal of α\alpha and β\beta?

Solution

Case I: Equation whose roots are α\alpha and β\beta ⇒ 2x2+3x−4=0 2x^{2} + 3x - 4 = 0

Equation whose roots are (α+5)(\alpha + 5) and (β+5)(\beta + 5) ⇒ 2(x−5)2+3(x−5)−4=0 2(x - 5)^{2} + 3(x - 5) - 4 = 0
⇒ 2x2−17x+31=0 2x^{2} - 17x + 31 = 0

Equation whose roots are (α+52)\left(\dfrac{\alpha + 5}{2} \right) and (β+52)\left(\dfrac{\beta + 5}{2} \right) ⇒ 2(2x)2−17(2x)+31=0 2(2x)^{2} - 17(2x) + 31 = 0

⇒ 8x2−34x+31=0 8x^{2} - 34x + 31 = 0

Case II: Equation whose roots are 1α\dfrac{1}{\alpha} and 1β\dfrac{1}{\beta} ⇒ −4x2+3x+2=0 -4x^{2} + 3x + 2 = 0

⇒ 4x2−3x−2=0 4x^{2} - 3x - 2 = 0

Equation whose roots are (23α)\left(\dfrac{2}{3\alpha} \right) and (23β)\left(\dfrac{2}{3\beta} \right) ⇒ 4(32×x)−2=0 4 \left(\dfrac{3}{2} \times x \right) - 2 = 0

⇒ 18x2−9x−4=0 18x^{2} - 9x - 4 = 0

Answer: (I) 8x2−34x+31=08x^{2} - 34x + 31 = 0; (II) 18x2−9x−4=018x^{2} - 9x - 4 = 0


Example 10

In the equation 4x2−12x+m=04x^{2} - 12x + m = 0, if one root is 55 times the other, then m=m = ?

Solution

Ratio of roots =p:q=1:5= p : q = 1 : 5

b2pq=ac(p+q)2b^{2}pq = ac(p + q)^{2}
⇒ (−12)2×1×5=4m(1+5)2 (-12)^{2} \times 1 \times 5 = 4m(1 + 5)^{2}
⇒ 144×5=4m×36 144 \times 5 = 4m \times 36
⇒ m=5 m = 5

Answer: 55


3.6 Applying Binomial Identities to find new Equation

Example 11

If α\alpha and β\beta are the roots of the equation x2+5x−3=0x^{2} + 5x - 3 = 0, what is the equation whose roots are
(I) 1α2\dfrac{1}{\alpha^{2}} and 1β2\dfrac{1}{\beta^{2}}?

(II) α2β\dfrac{\alpha^{2}}{\beta} and β2α\dfrac{\beta^{2}}{\alpha}?

Solution

Sum of roots =α+β=−5= \alpha + \beta = -5
Product of roots =αβ=−3= \alpha \beta = -3

Case I: Where roots are 1α2\dfrac{1}{\alpha^{2}} and 1β2\dfrac{1}{\beta^{2}}

Sum of roots =1α2+1β2=α2+β2(α2β2)= \dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} = \dfrac{\alpha^{2} + \beta^{2}}{(\alpha^{2} \beta^{2})} =(α+β)2−2αβ(αβ)2=(−5)2−2×−3(−3)2= \dfrac{(\alpha + \beta)^{2} - 2 \alpha \beta}{(\alpha \beta)^{2}} = \dfrac{(-5)^{2} - 2 \times -3}{(-3)^{2}}

=319= \dfrac{31}{9}

Product of roots =1α2×1β2=1(αβ)2=19= \dfrac{1}{\alpha^{2}} \times \dfrac{1}{\beta^{2}} = \dfrac{1}{(\alpha \beta)^{2}} = \dfrac{1}{9}

New Equation ⇒ x2−319x+19=0 x^{2} - \dfrac{31}{9}x + \dfrac{1}{9} = 0 ⇒ 9x2−31x+1=0 9x^{2} - 31x + 1 = 0

Case II: Where roots are α2β\dfrac{\alpha^{2}}{\beta} and β2α\dfrac{\beta^{2}}{\alpha}

Sum of roots =α2β+β2α=α3+β3αβ= \dfrac{\alpha^{2}}{\beta} + \dfrac{\beta^{2}}{\alpha} = \dfrac{\alpha^{3} + \beta^{3}}{\alpha \beta} =(α+β)3−3αβ(α+β)αβ= \dfrac{(\alpha + \beta)^{3} - 3 \alpha \beta (\alpha + \beta)}{\alpha \beta} =(−5)3−3×−3×−5−3=1703= \dfrac{(-5)^{3} - 3 \times -3 \times -5}{-3} = \dfrac{170}{3}

Product of roots =α2β×β2α=αβ=−3= \dfrac{\alpha^{2}}{\beta} \times \dfrac{\beta^{2}}{\alpha} = \alpha \beta = -3

New Equation ⇒ x2−1703x−3=0 x^{2} - \dfrac{170}{3}x - 3 = 0 ⇒ 3x2−170x−9=0 3x^{2} - 170x - 9 = 0

Answer: (I) 9x2−31x+1=09x^{2} - 31x + 1 = 0; (II) 3x2−170x−9=03x^{2} - 170x - 9 = 0


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