Question

prove that............. don't spam​

Answer 1

We have 2 Eqns. and the 2nd have common roots. Therefore the discriminant of the 2nd Eqn. is 0.

${x}^{2} - ax + b = 0 \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{a + - \sqrt{ {a}^{2} - 4b } }{2} \\ x = \frac{a}{2}$

Now, Equating Discriminant to be 0.

${a}^{2} - 4b = 0 \\ {a}^{2} = 4b$

From the value of x we get

$x = \frac{a}{2} \\ sqaring \: bs \\ {x}^{2} = \frac{ {a}^{2} }{4} \\ {x}^{2} = \frac{4b}{4} \\ {x}^{2} = b$

Putting the value of x² in 1st Eqn.

${x}^{2} - px + q = 0 \\ b - px + q = 0 \\ b + q = px \\ b + q = p( \frac{a}{2} ) \\ b + q = \frac{ap}{2}$

HENCE PROVED.

9TH GRADER.

MARK AS BRAINLIEST

Answer 2

Answer:

We have 2 Eqns. and the 2nd have common roots. Therefore the discriminant of the 2nd Eqn. is 0.

\begin{lgathered}{x}^{2} - ax + b = 0 \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{a + - \sqrt{ {a}^{2} - 4b } }{2} \\ x = \frac{a}{2}\end{lgathered}

x

2

−ax+b=0

x=

2a

−b+−

b

2

−4ac

x=

2

a+−

a

2

−4b

x=

2

a

Now, Equating Discriminant to be 0.

\begin{lgathered}{a}^{2} - 4b = 0 \\ {a}^{2} = 4b\end{lgathered}

a

2

−4b=0

a

2

=4b

From the value of x we get

\begin{lgathered}x = \frac{a}{2} \\ sqaring \: bs \\ {x}^{2} = \frac{ {a}^{2} }{4} \\ {x}^{2} = \frac{4b}{4} \\ {x}^{2} = b\end{lgathered}

x=

2

a

sqaringbs

x

2

=

4

a

2

x

2

=

4

4b

x

2

=b

Putting the value of x² in 1st Eqn.

\begin{lgathered}{x}^{2} - px + q = 0 \\ b - px + q = 0 \\ b + q = px \\ b + q = p( \frac{a}{2} ) \\ b + q = \frac{ap}{2}\end{lgathered}

x

2

−px+q=0

b−px+q=0

b+q=px

b+q=p(

2

a

)

b+q=

2

ap

HENCE PROVED.

9TH GRADER.

MARK AS BRAINLIEST