Class 10 Math All topics Quadratic Equations

Find the value 'p' so that the equation $4x_{2}−8px+9=0$ has roots whose difference is 4.

Solution:

We know that if $m$ and $n$ are the roots of a quadratic equation $ax_{2}+bx+c=0$, the sum of the roots is $m+n=−ab $ and the product of the roots is $mn=ac $.

Let $m$ and $n$ be the roots of the given quadratic equation $4x_{2}−8px+9=0$. It is given that the difference of the roots is $4$, therefore,

$m−n=4........(1)$

The equation $4x_{2}−8px+9=0$ is in the form $ax_{2}+bx+c=0$ where $a=4,b=−8q$ and $c=9$.

The sum of the roots is:

$m+n=−ab =−4(−8q) =2q....(2)$

The product of the roots is $ac $ that is:

$mn=ac =49 ......(3)$

Now, we know the identity $(m+n)_{2}=(m−n)_{2}+4mn$, therefore, using equations 1,2 and 3, we have

$(m+n)_{2}=(m−n)_{2}+4mn⇒(2p)_{2}=4_{2}+(4×49 )⇒4p_{2}=16+9⇒4p_{2}=25⇒p_{2}=425 ⇒p=±425 ⇒p=±25 $

Hence, the value of $p=±25 $.

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introduction to trigonometry

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