Name: Condition for Common Roots of Quadratic Equation
Uploaded: 2019-07-19
Duration: 36 min 41 s
Description: In this video, we will learn the condition for one common root and the condition for both roots common for two given quadratic equations with the help of exa...

Question

Let

a x^{4} + b x^{3} + x^{2} + (3 - a) x + 3 = 0

and

x^{2} + (2 - a) x + 3 = 0

have common roots. If

a \in (- 1, 5)

and

b \in R

, then find

| a + 12 b |

Accepted Answer

$x^{2} + (2 - a) x + 3 = 0$

Discriminant of above quadratic equation, $D = {(2 - a)}^{2} - 4 (1) (3)$

$\Rightarrow D = a^{2} + 4 - 4 a - 12$

$\Rightarrow D = a^{2} - 4 a - 8$

$\Rightarrow D = \{a - (2 + 2 \sqrt{3})\} \{a - (2 - 2 \sqrt{3})\}$

{ $∵$ Roots of quadratic equation $p x^{2} + q x + r = 0$ is given by $x = \frac{- q \pm \sqrt{q^{2} - 4 p r}}{2 p}$ }

$a x^{4} + b x^{3} + x^{2} + (3 - a) x + 3 = 0$

$\Rightarrow$ Value of $a$ lies between the roots of quadratic expression $a^{2} - 4 a - 8$

As we know $f (x) = p x^{2} + q x + r < 0; x \in (α, β)$ , where $α$ and $β$ are the roots of quadratic equation

$\Rightarrow$ For all $a \in (- 1, 5), D < 0$ .

So, Equation has imaginary roots and imaginary roots occur as conjugate pairs, so both roots will be common. It also means that quadratic will be factor of biquadratic.

We can write biquadratic as a multiple of two quadratic factors.

$a x^{4} + b x^{3} + x^{2} + (3 - a) x + 3 = (x^{2} + (2 - a) x + 3) (a_{1} x^{2} + b_{1} x + c_{1})$ .

Comparing coefficient of $x^{4} \Rightarrow a = a_{1}$ ... $(i)$

Comparing constant term $\Rightarrow 3 = 3 c_{1} \Rightarrow c_{1} = 1$ ... $(ii)$

Comparing coefficient of $x \Rightarrow 3 b_{1} + (2 - a) c_{1} = 3 - a$

$3 b_{1} + (2 - a) = 3 - a \Rightarrow b_{1} = \frac{1}{3}$ ... $(iii)$

Replacing values of $a_{1}, b_{1} & c_{1}$ from $(i), (ii) & (iii)$

$a x^{4} + b x^{3} + x^{2} + (3 - a) x + 3 = (x^{2} + (2 - a) x + 3) (a x^{2} + \frac{1}{3} x + 1)$

Comparing coefficient of $x^{2} \Rightarrow 3 a + 1 + (2 - a) \frac{1}{3} = 1$

$\Rightarrow 3 a + \frac{(2 - a)}{3} = 0$

$\Rightarrow 9 a + 2 - a = 0$

$\Rightarrow a = - \frac{1}{4}$ ... $(iv)$

Comparing coefficient of $x^{3} \Rightarrow \frac{1}{3} + a (2 - a) = b$

$\Rightarrow b = \frac{1}{3} + (- \frac{1}{4}) (2 - (- \frac{1}{4})) = - \frac{11}{48}$ ... $(v)$

We have to find value of $|a + 12 b|$

Replacing values of $a & b$ form $(iv)$ and $(v)$

$\Rightarrow |a + 12 b| = |- \frac{1}{4} + 12 (- \frac{11}{48})| = |- \frac{1}{4} - \frac{11}{4}| = 3$

Let $a x^{4} + b x^{3} + x^{2} + (3 - a) x + 3 = 0$ and $x^{2} + (2 - a) x + 3 = 0$ have common roots. If $a \in (- 1, 5)$ and $b \in R$ , then find $| a + 12 b |$

Important Questions on Quadratic Equations

Learn and Prepare for any exam you want !

Let ax4+bx3+x2+3-ax+3=0 and x2+2-ax+3=0 have common roots. If a∈(-1,5) and b∈R, then find |a+12 b|

Important Questions on Quadratic Equations

Learn and Prepare for any exam you want !

Let $a x^{4} + b x^{3} + x^{2} + (3 - a) x + 3 = 0$ and $x^{2} + (2 - a) x + 3 = 0$ have common roots. If $a \in (- 1, 5)$ and $b \in R$ , then find $| a + 12 b |$