If the quadratic equations ax 2-2 cx - b -0 and ax 2-2 bx - c -0- b - c have a common root- then a -4 b -4 c equals to

Let α be the common root. ⇒ aα^2+2cα+b=0 and aα^2+2bα+c=0 ⇒ 2α(c b)+(b c)=0 ⇒α=12 (∵ b c) So, a(12)^2+2c(12)+b=0 ⇒ a+4b+4c=0 ...

You visited us 50 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Common Roots

If the quadra...

Question

If the quadratic equations $a x^{2} + 2 c x + b = 0$ and $a x^{2} + 2 b x + c = 0, (b \neq c)$ have a common root, then $a + 4 b + 4 c$ equals to

Open in App

Solution

Let $α$ be the common root.
$\Rightarrow a α^{2} + 2 c α + b = 0$ and $a α^{2} + 2 b α + c = 0$
$\Rightarrow 2 α (c - b) + (b - c) = 0$
$\Rightarrow α = \frac{1}{2} (∵ b \neq c)$

So, $a {(\frac{1}{2})}^{2} + 2 c (\frac{1}{2}) + b = 0$
$\Rightarrow a + 4 b + 4 c = 0$

Suggest Corrections

Similar questions

Q. If the quadratic equation,

a x^{2} + 2 c x + b = 0

a x^{2} + 2 b x - 5 x + c = 0 (b \neq c)

hava a common root then find the value of

a + 4 b + 4 c

If equation $a x^{2} + 2 c x + b = 0$ and $a x^{2} + 2 b x + c = 0$ have one root in common, then a + 4b + 4c equals

Q. If the quadratic equations

a x^{2} + 2 c x + b = 0

and

a x^{2} + 2 b x + c = 0, (b \neq c)

have a common root, then the value of

a + 4 b + 4 c

Q. If the equation

a x^{2} + 2 b x + c = 0

and

a x^{2} + 2 c x + b = 0, a \neq 0, b \neq c

, have a common root, then their other roots are the roots of the quadratic equation

Suppose that the three quadratic equations $a x^{2} - 2 b x + c = 0, b x^{2} - 2 c x + a = 0$ and $c x^{2} - 2 a x + b = 0$ all have only positive roots. Then,

Join BYJU'S Learning Program

If the quadratic equations ax2+2cx+b=0 and ax2+2bx+c=0, (b≠c) have a common root, then a+4b+4c equals to

Let α be the common root. ⇒aα2+2cα+b=0 and aα2+2bα+c=0 ⇒2α(c−b)+(b−c)=0 ⇒α=12 (∵b≠c) So, a(12)2+2c(12)+b=0 ⇒a+4b+4c=0

If the quadratic equations $a x^{2} + 2 c x + b = 0$ and $a x^{2} + 2 b x + c = 0, (b \neq c)$ have a common root, then $a + 4 b + 4 c$ equals to

Let $α$ be the common root.
$\Rightarrow a α^{2} + 2 c α + b = 0$ and $a α^{2} + 2 b α + c = 0$
$\Rightarrow 2 α (c - b) + (b - c) = 0$
$\Rightarrow α = \frac{1}{2} (∵ b \neq c)$

So, $a {(\frac{1}{2})}^{2} + 2 c (\frac{1}{2}) + b = 0$
$\Rightarrow a + 4 b + 4 c = 0$