Question

FIND THE VALUE OF P, FOR WHICH ONE ROT OF THE QUADRATIC EQUATION px^2-14x+8 is 6 times the other

Answer 1

Answer:

4

Step-by-step explanation:

roots be a , 6a

7a = 14/p

6a² = 8/p

6a/7 = 3/7

a = 1/2 , 6a = 3

p = 14/7a = 2/a = 4

Answer 2

Step-by-step explanation:

Given : A quadratic equation $px^{2} -14x+8$ whose first root is  $6$ times other.

To find : The value of p.

Formula used : If we have standard quadratic equation then sum and product of roots are,

Sum of roots $=\frac{-b}{a}$                          

Product of roots $=\frac{c}{a}$

a is the coefficient of $x^{2}$ , b is the coefficient of $x$  and c is the constant.

• Calculation for roots

First of all Let us take two roots of given quadratic equation by using given data,

First root  $=y$

Second root  $=6y$

So the sum of roots $=y+6y$                   and       product of roots $=y*6y$

                                $=7y$                                                                    $=6y^{2}$    

we have quadratic equation and we can equate it with zero,

⇒  $px^{2} -14x+8=0$

Now the sum and products of roots by using coefficients,

Sum of roots $=\frac{-b}{a}$                             Product of roots $=\frac{c}{a}$

                      $=\frac{-(-14)}{p}$                                                  $=\frac{8}{p}$

                      $=\frac{14}{p}$                                                        $=\frac{8}{p}$

As we know sum and product of roots by coefficients and roots should be same so we can equate them as,

Sum of roots $=\frac{14}{p}$                               Product of roots  $=\frac{8}{p}$

        $7y= \frac{14}{p}$                                                        $6y^{2} =\frac{8}{p}$

taking  $\frac{1}{p}$   remain at R.H.S and other terms to L.H.S in both sum and product of roots,

          $\frac{7y}{14} =\frac{1}{p}$                                                            $\frac{6y^{2} }{8} =\frac{1}{p}$

          $\frac{y}{2} =\frac{1}{p}$      --(1)                                                   $\frac{3y^{2} }{4} =\frac{1}{p}$   --(2)

R.H.S of both equations (1) and (2) are same hence the L.H.S of equations are also same so by equating equations (1) and (2),

⇒  $\frac{y}{2} =\frac{3y^{2} }{4}$

taking similar terms at same side,

⇒  $\frac{4}{2} =\frac{3y^{2} }{y}$

⇒  $2=3y$

⇒  $y=\frac{2}{3}$

we get the first root is $\frac{2}{3}$ and

second root $=6*y$

                     $=6*\frac{2}{3}$

                     $=4$

So the roots of equation are $\frac{2}{3},4$.

• Calculation for p,

We have the value of y is

⇒  $y=\frac{2}{3}$

by putting this y in equation (1) we get the value of p,

⇒  $\frac{\frac{2}{3} }{2} =\frac{1}{p}$  

⇒  $\frac{2 }{2*3} =\frac{1}{p}$

⇒ $\frac{1 }{3} =\frac{1}{p}$

doing cross multiplication,

⇒  $p=3$

we get the value of p is 3.

So the quadratic equation by putting the value of p, is,

⇒  $3x^{2} -14x+8$.