Question

13. Find the points on X-axis which are at a distance of 2V5 units from the
point (7,-4)? *
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Answer 1

The answer to this question is

2√​5​​​=√​(7−x)​2​​+(−4−0)​2​​​​​​=>(2√​5​​​)​2​​=(7−x)​2​​+16​=>20=49+x​2​​−14x+16​=>x​2​​−14x+45=0​=>x​2​​−9x−5x+45=0​=>x(x−9)−5(x−9)​=>(x−5)(x−9)=0​=>x=9andx=5​​  

there are two points i.e., (5, 0) and (9, 0).

Hope dis helps!!

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Answer 2

(5,0)&(9,0)

(2√5)²=20

4²=16

20-16=x²

x²=4

x=±2

(7±2,0)

(5,0)&(9.0)