Answer:
Step-by-step explanation:
Answer:
e e enene
Step-by-step explanation:
enenenee
Answer:
and ![\frac{-6-\sqrt{78} }{6}](https://tex.z-dn.net/?f=%5Cfrac%7B-6-%5Csqrt%7B78%7D%20%7D%7B6%7D)
Step-by-step explanation:
We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.
So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)
Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.
This formula gives if, ax² +bx +c =0, the the two roots of the equation are
![\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%2B%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D%20%2C%20%5Cfrac%7B-b-%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D)
Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be
![\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6}, \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}](https://tex.z-dn.net/?f=%5Cfrac%7B-12%2B%5Csqrt%7B12%5E%7B2%7D-4%2A6%2A%28-7%29%20%7D%20%7D%7B2%2A6%7D%2C%20%20%5Cfrac%7B-12-%5Csqrt%7B12%5E%7B2%7D-4%2A6%2A%28-7%29%20%7D%20%7D%7B2%2A6%7D)
= ![\frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}](https://tex.z-dn.net/?f=%5Cfrac%7B-12%2B%5Csqrt%7B312%7D%20%7D%7B12%7D%2C%5Cfrac%7B-12-%5Csqrt%7B312%7D%20%7D%7B12%7D)
=
and ![\frac{-6-\sqrt{78} }{6}](https://tex.z-dn.net/?f=%5Cfrac%7B-6-%5Csqrt%7B78%7D%20%7D%7B6%7D)
Hence, x= ![\frac{-6+\sqrt{78} }{6}](https://tex.z-dn.net/?f=%5Cfrac%7B-6%2B%5Csqrt%7B78%7D%20%7D%7B6%7D)
and x= ![\frac{-6-\sqrt{78} }{6}](https://tex.z-dn.net/?f=%5Cfrac%7B-6-%5Csqrt%7B78%7D%20%7D%7B6%7D)
(Answer)
Answer:
A. segment PS ≅ segment QR
Step-by-step explanation:
Given that the quadrilateral PQRS is a rectangle then;
- The diagonals are PR and SQ
- Point of intersection of diagonals is T
- Segment PS≅ Segment QR
- ∠PSQ≅ ∠SQR
- segment SR≅ segment PQ