Question

Check whether the following equations has real roots, 9x2

-15x+6=0 and if it has, find them.​

Answer 1

Answer:

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Answer 2

Answer:

$\large x=1 \ or \ x=\dfrac{2}{3}$

Step-by-step explanation:

Given :

$\large p(x)=9x^2-15x+6=0$

We know condition for real    D ≥ 0

where $\large D=b^2-4ac$

Comparing values from given

a = 9 , b = - 15 and c = 6

Now putting values here we get

$\large (-15)^2-4\times9\times6\geq0\\\\\\\large 225-216 \geq0\\\\\\\large 9\geq0$

Since condition is satisfied.

P ( x ) has a real roots.

By splitting term mid method we get

$\large p(x)=9x^2-15x+6=0\\\\\\\large p(x)=9x^2-9x-6x+6=0\\\\\\\large p(x)=9x(x-1)-6(x-1)=0\\\\\\\large p(x)=(x-1)(9x-6)=0\\\\\\\large (x-1)=0 \ or \ (9x-6)=0\\\\\\\large x=1 \ or \ 9x=6\\\\\\\large x=1 \ or \ x=\dfrac{2}{3}$

Thus we get roots also.