Question

Find the value of p for which the quadratic equation \[\left( {2p + 1} \right){x^2} - \left( {7p + 2} \right)x + \left( {7p - 3} \right) = 0\] has equal roots. Also find the roots.

Answer 1

Hint: Given is a quadratic equation. But it is given with two variables such that when we will find the value of p we will get it in the form of \[a{x^2} + bx + c = 0\] . Also, the condition for the roots is that they are equal. So we will use the discriminant and quadratic equation formula to find the value of p and the roots as well.

Complete step by step solution:
Given the equation is,
 \[\left( {2p + 1} \right){x^2} - \left( {7p + 2} \right)x + \left( {7p - 3} \right) = 0\]
Comparing with general quadratic equation we get,
 \[a = 2p + 1,b = - \left( {7p + 2} \right),c = 7p - 3\]
We know that the value of discriminant \[{b^2} - 4ac\] decides the nature of the roots. We are given that the roots are equal.
So we can say that \[{b^2} - 4ac = 0\]
 \[{b^2} = 4ac\]
Now putting the values we get,
 \[{\left( { - \left( {7p + 2} \right)} \right)^2} = 4 \times \left( {2p + 1} \right) \times \left( {7p - 3} \right)\]
On rearranging we get,
 \[{\left( {7p + 2} \right)^2} = 4\left[ {\left( {2p + 1} \right)\left( {7p - 3} \right)} \right] \]
Taking the square on LHS and multiplying the brackets on RHS we get,
 \[49{p^2} + 2 \times 7p \times 2 + {2^2} = 4\left[ {2p\left( {7p - 3} \right) + 1\left( {7p - 3} \right)} \right] \]
Now multiplying the terms on RHS we get,
 \[49{p^2} + 28p + 4 = 4\left[ {2p \times 7p - 2p \times 3 + 7p - 3} \right] \]
On multiplying we get,
 \[49{p^2} + 28p + 4 = 4\left[ {14{p^2} - 6p + 7p - 3} \right] \]
On adding the same terms,
 \[49{p^2} + 28p + 4 = 4\left[ {14{p^2} + p - 3} \right] \]
On multiplying by 4 we get,
 \[49{p^2} + 28p + 4 = 56{p^2} + 4p - 12\]
On rearranging the terms we get,
 \[56{p^2} - 49{p^2} + 4p - 28p - 12 - 4 = 0\]
Performing the operations on same terms we get,
 \[7{p^2} - 24p - 16 = 0\]
Now we again get a quadratic equation. We will solve this to get the values of p.
Using quadratic equation we get,
 \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - \left( { - 24} \right) \pm \sqrt {{{24}^2} - 4 \times 7 \times \left( { - 16} \right)} }}{{2 \times 7}}\]
On calculating the roots and brackets,
 \[ = \dfrac{{24 \pm \sqrt {576 + 448} }}{{14}}\]
On adding the numbers in the root,
 \[ = \dfrac{{24 \pm \sqrt {1024} }}{{14}}\]
We know that 1024 is the perfect square of 32.
 \[ = \dfrac{{24 \pm 32}}{{14}}\]
Now the values of p can be drawn as,
 \[p \Rightarrow \dfrac{{24 + 32}}{{14}}or\dfrac{{24 - 32}}{{14}}\]
On performing respective operations,
 \[p \Rightarrow \dfrac{{56}}{{14}}or\dfrac{{ - 8}}{{14}}\]
On simplifying the ratios we get,
 \[p \Rightarrow 4or\dfrac{{ - 4}}{7}\]
So these are the values of p. using these values one by one we will get the values of the roots.

For p=4:
 \[\left( {2 \times 4 + 1} \right){x^2} - \left( {7 \times 4 + 2} \right)x + \left( {7 \times 4 - 3} \right) = 0\]
On multiplying the numbers we get,
 \[9{x^2} - 30x + 25 = 0\]
This is again a quadratic equation. We will again use the quadratic formula to find the answer.
 \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{30 \pm \sqrt {{{30}^2} - 4 \times 9 \times 25} }}{{2 \times 9}}\]
On calculating we get,
 \[ = \dfrac{{30 \pm \sqrt {900 - 900} }}{{18}}\]
Thus we get,
 \[ = \dfrac{{30}}{{18}}\]
On simplifying we get,
 \[x = \dfrac{5}{3}\]
This is the value of the roots if p=4.

For \[p = \dfrac{{ - 4}}{7}\] :
 \[\left( {2 \times \dfrac{{ - 4}}{7} + 1} \right){x^2} - \left( {7 \times \dfrac{{ - 4}}{7} + 2} \right)x + \left( {7 \times \dfrac{{ - 4}}{7} - 3} \right) = 0\]
Taking the LCM in first bracket and cancelling 7 in other two brackets,
 \[\left( {\dfrac{{ - 8 + 7}}{7}} \right){x^2} - \left( { - 4 + 2} \right)x + \left( { - 4 - 3} \right) = 0\]
On calculating we get,
 \[\dfrac{{ - 1}}{7}{x^2} + 2x - 7 = 0\]
Multiplying both sides by 7,
 \[ - {x^2} + 14x - 49 = 0\]
On multiplying by -1 we get,
 \[{x^2} - 14x + 49 = 0\]
If we observe that this is the perfect square of \[{\left( {x - 7} \right)^2} = 0\]
So taking the roots we get,
 \[x - 7 = 0\]
Thus the value of x will be,
 \[x = 7\]
Thus the values of x are \[x = \dfrac{5}{3}\] and x=7.

Note: Here note that we have used quadratic equation formulas so many times. So don’t get confused. First we found the values of p so that we can form the quadratic equation. Then we actually found the roots using the values of p. so the answer id values of p and the roots are values of x. thus values of x are same.