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CAT 2025 Lesson : Quadratic Equations - Higher Order Equations

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5. Higher Order Equations

An equation of the form is a0xn+a1xn−1+a2xn−2+....+an−2x2+an−1x1+an=0a_{0}x^{n} + a_{1}x^{n - 1} + a_{2}x^{n - 2} + .... + a_{n -2}x^{2} + a_{n - 1}x^{1} + a_{n} = 0 called a nthn^{\text{th}} degree equation where a0,a1,a3,....,an−1,ana_{0}, a_{1}, a_{3}, ...., a_{n - 1}, a_{n} are real numbers . A nthn^{\text{th}} degree equation will have nn roots.

5.1 Quadratic Equation

We have learnt this in the earlier section.

If α\alpha and β\beta are the roots of an equation, then
(x−α)(x−β)=0(x - \alpha)(x - \beta) = 0
⇒ x2−(α+β)x+(αβ)=0⟶(1) x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0 \longrightarrow (1)

General form of quadratic equation is ax2+bx+c=0ax^{2} + bx + c = 0

⇒ x2+bax+ca=0⟶(2) x^{2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0 \longrightarrow (2)

Comparing coefficients of variables and the constant in (1)(1) and (2)(2), we note the following.

α+β=−ba\alpha + \beta = \dfrac{-b}{a} αβ=ca\alpha \beta = \dfrac{c}{a}


The above method has been applied to form the identities for cubic and biquadratic equations.

5.2 Cubic Equation

If α,β\alpha, \beta and γ\gamma are the roots of the equation ax3+bx2+cx+d=0ax^{3} + bx^{2} + cx + d = 0, then

α+β+γ=−ba\alpha + \beta + \gamma = \dfrac{-b}{a} αβ+βγ+γα=ca\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} αβγ=−da\alpha \beta \gamma = \dfrac{-d}{a}


5.3 Bi-quadratic Equation

If α,β,γ\alpha, \beta, \gamma and δ\delta are the roots of the equation ax4+bx3+cx2+dx+e=0ax^{4} + bx^{3} + cx^{2} + dx + e = 0,

α+β+γ+δ=−ba\alpha + \beta + \gamma + \delta = \dfrac{-b}{a} αβ+αγ+αδ+βγ+βδ+γδ=ca\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \dfrac{c}{a} αβγ+αβδ+αγδ+βγδ=−da\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = \dfrac{-d}{a} αβγδ=ea\alpha \beta \gamma \delta = \dfrac{e}{a}


Example 23

If p,qp, q and rr are the roots of the equation 2x3−3x2−23x+12=02x^{3} - 3x^{2} - 23x + 12 = 0, then 1pq+1qr+1rp=\dfrac{1}{pq} + \dfrac{1}{qr} + \dfrac{1}{rp} = ?

(1) −0.25-0.25            (2) −0.5-0.5            (3) 0.250.25            (4) 0.50.5           

Solution

1pq+1qr+1rp=p+q+rpqr\dfrac{1}{pq} + \dfrac{1}{qr} + \dfrac{1}{rp} = \dfrac{p + q + r}{pqr}

In the cubic equation,
Sum of roots =p+q+r=−ba=32= p + q + r = \dfrac{-b}{a} = \dfrac{3}{2}

Product of roots =pqr=−da=−122=−6= pqr = \dfrac{-d}{a} = \dfrac{-12}{2} = -6

∴ p+q+rpqr=32×1−6=−0.25\dfrac{p + q + r}{pqr} = \dfrac{3}{2} \times \dfrac{1}{-6} = -0.25

Answer: (1) −0.25(1) \space -0.25


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